#include iostream #include fstream #include shirtType.h.pdf
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#include #include #include \"shirtType.h\" using namespace std; void getInput(shirtType s[],ifstream& infile, int& numShirts); int main() { int numShirts; shirtType shirts[500]; ifstream infile; infile.open(\"h:\\\\shirtlab.txt\"); getInput(shirts,infile,numShirts); for(int i =0; i>id; infile.ignore(100,\'\ \'); getline(infile,c); infile>>size>>p>>q; while(infile) { s[index].setAll(id, c, size, p, q); index++; infile>>id; infile.ignore(100,\'\ \'); getline(infile,c); infile>>size>>p>>q; } numShirts=index; } See also the description of the class shirtType below: #include #include using namespace std; class shirtType { private: int shirtID; string color; string size; double price; int quantity; public: void setAll(int i,string c,string s, double p, int q); void print() const; void newQuan(); shirtType(); shirtType(int,string,string,double,int); }; shirtType::shirtType(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } shirtType::shirtType() { shirtID=0; color=\"\"; size=\"\"; price=0.0; quantity=0; } void shirtType::setAll(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } void shirtType::newQuan() { cout<<\"Enter the new quantity of shirts: \"<>quantity; } void shirtType::print() const { cout<<\"Shirt ID: \"< Solution //shirtType.h #include #include using namespace std; class shirtType { private: int shirtID; string color; string size; double price; int quantity; public: void setAll(int i,string c,string s, double p, int q); void print() const; void newQuan(); int changeQuan(); int getShirtId(int); shirtType(); int search(int); shirtType(int,string,string,double,int); }; int shirtType::search(int sellId) { if(shirtID == sellId) return 1; else return 0; } int shirtType::getShirtId(int i) { return shirtID; } shirtType::shirtType(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } shirtType::shirtType() { shirtID=0; color=\"\"; size=\"\"; price=0.0; quantity=0; } void shirtType::setAll(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } void shirtType::newQuan() { int q; cout<<\"Enter the new quantity of shirts: \"<>q; quantity += q; } int shirtType::changeQuan() { if(quantity > 0 ) { quantity -= 1; return 1; } { return 2; } } void shirtType::print() const { cout<<\"Shirt ID: \"< #include #include #include \"shirtType.h\" #include using namespace std; void getInput(shirtType s[],ifstream& infile, int& numShirts); void addShirt(shirtType s[], int& numShirts); void sellShirt(shirtType s[], int& numShirts); void addstockShirt(shirtType s[], int&.
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![#include
#include
#include "shirtType.h"
using namespace std;
void getInput(shirtType s[],ifstream& infile, int& numShirts);
int main()
{
int numShirts;
shirtType shirts[500];
ifstream infile;
infile.open("h:shirtlab.txt");
getInput(shirts,infile,numShirts);
for(int i =0; i>id;
infile.ignore(100,' ');
getline(infile,c);
infile>>size>>p>>q;
while(infile)
{
s[index].setAll(id, c, size, p, q);
index++;
infile>>id;
infile.ignore(100,' ');
getline(infile,c);
infile>>size>>p>>q;
}
numShirts=index;
}
See also the description of the class shirtType below:
#include
#include
using namespace std;
class shirtType
{
private:](https://image.slidesharecdn.com/includeiostreamincludefstreamincludeshirttype-230702074645-dab09897/85/include-iostream-include-fstream-include-shirtType-h-pdf-1-320.jpg)
Recommended
C++ PROGRAMThis program builds on the code below#include iostr.pdf
C++ PROGRAM
This program builds on the code below:
#include
#include
#include \"shirtType.h\"
using namespace std;
void getInput(shirtType s[],ifstream& infile, int& numShirts);
int main()
{
int numShirts;
shirtType shirts[500];
ifstream infile;
infile.open(\"h:\\\\shirtlab.txt\");
getInput(shirts,infile,numShirts);
for(int i =0; i>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
while(infile)
{
s[index].setAll(id, c, size, p, q);
index++;
infile>>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
}
numShirts=index;
}
See also the description of the class shirtType below:
#include
#include
using namespace std;
class shirtType
{
private:
int shirtID;
string color;
string size;
double price;
int quantity;
public:
void setAll(int i,string c,string s, double p, int q);
void print() const;
void newQuan();
shirtType();
shirtType(int,string,string,double,int);
};
shirtType::shirtType(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
shirtType::shirtType()
{
shirtID=0;
color=\"\";
size=\"\";
price=0.0;
quantity=0;
}
void shirtType::setAll(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
void shirtType::newQuan()
{
cout<<\"Enter the new quantity of shirts: \"<>quantity;
}
void shirtType::print() const
{
cout<<\"Shirt ID: \"<
Solution
#include
#include
struct node
{
int data;
struct node *next;
}*head;
void append(int nkkm)
{
struct node *temp,*right;
temp= (struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
right=(struct node *)head;
while(right->next != NULL)
right=right->next;
right->next =temp;
right=temp;
right->next=NULL;
}
void add( int nkkm )
{
struct node *temp;
temp=(struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
if (head== NULL)
{
head=temp;
head->next=NULL;
}
else
{
temp->next=head;
head=temp;
}
}
void addafter(int nkkm, int loc)
{
int i;
struct node *temp,*left,*right;
right=head;
for(i=1;inext;
}
temp=(struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
left->next=temp;
left=temp;
left->next=right;
return;
}
void addaShirt(int nkkm)
{
int c=0;
struct node *temp;
temp=head;
if(temp==NULL)
{
add(nkkm);
}
else
{
while(temp!=NULL)
{
if(temp->datanext;
}
if(c==0)
add(nkkm);
else if(cdata==nkkm)
{
if(temp==head)
{
head=temp->next;
free(temp);
return 1;
}
else
{
prev->next=temp->next;
free(temp);
return 1;
}
}
else
{
prev=temp;
temp= temp->next;
}
}
return 0;
}
void display(struct node *r)
{
r=head;
if(r==NULL)
{
return;
}
while(r!=NULL)
{
printf(\"%d \",r->data);
r=r->next;
}
printf(\"\ \");
}
int count()
{
struct node *n;
int c=0;
n=head;
while(n!=NULL)
{
n=n->next;
c++;
}
return c;
}
int main()
{
int i,nkkm;
struct node *n;
head=NULL;
while(1)
{
printf(\"1.Add a shirt\ \");
printf(\"2.Sell a shirt\ \");
printf(\"3.Res.
Basic C Programming Lab Practice
In this slide, we discussed the assignment of solving known programming problems.
Programming Fundamentals Arrays and Strings
Programming Fundamentals
Arrays and Strings
• Arrays
• Initializing arrays
• Multidimensional arrays
• Arrays as arguments to functions
• Strings
• String functions
Accessing Individual Components
Declaring Arrays
Arrays: Example Garbage
Multidimensional Arrays
1) Write a C program using arrays that produces the multiplication of two 2x2 matrices.
3-Dimensional Array
Write a program of your own choice that makes use of arrays of more than 2 dimensions.
2-Dimensional Arrays
Twitter sentiment analysis for cryptoassets
python code
Please email at: mamoon@quake.vc or mik279@nyu.edu for questions about how to make modifications.
primary code credit: @ZeroVariance twitter handle.
#include stdio.h#include stdlib.h#include string.h#inclu.pdf
#include
#include
#include
#include
/*Structure for storing words received from each thread*/
struct return_val{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*Aggregates Frequency Sum of different threads*/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/*Function executed by each thread on separate line*/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] = \"p\' \'file.txt\'\\0\";
char command[100];
command[0] = \'\\0\';
strcat(command,cmd_p1);
strcat(command,cmd_p2);
strcat(command,cmd_p3);
//usleep(line_number);
char cmd[100] = \" | tr [:space:] \'\\\ \' | grep -v \'^\\\\s*$\' | sort | uniq -c | sort\\0\";
strcat(command,cmd);
FILE *in;
in= popen(command, \"r\");
rewind(in);
char buff[50];
int counter = 0;
while(fgets(buff,sizeof(buff),in))
{
char c=\' \';
int i = 0;
int cnt = atoi(buff);
arr[line_number-1].count[counter] = cnt;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+6];
i++;
}
int cnnt = 0;
while(c!=\' \')
{
c = buff[cnnt];
cnnt++;
}
i=0;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+cnnt];
i++;
}
sprintf(arr[line_number-1].wordlist[counter],\"%s\",buff);
counter++;
}
arr[line_number-1].count[counter] = -1;
fclose(in);
return NULL;
}
int main(void)
{
FILE *fp;
fp = fopen(\"file.txt\",\"r\");
char a[200];
lines = 0;
void *status = NULL;
while(fgets(a,sizeof(a),fp) !=NULL)
{
lines++;
}
fclose(fp);
arr = (struct return_val*) calloc(lines, sizeof(struct return_val) * lines);
threads = malloc(sizeof(pthread_t)*lines);
int i;
for(i=0;i= 0)
{
if(arr[i].count[ctr] > 0)
{
strcpy(final[final_ctr],arr[i].wordlist[ctr]);
final_cnt[final_ctr] = arr[i].count[ctr];
arr[i].count[ctr] = 0;
int sum = aggregate(arr[i].wordlist[ctr]);
final_cnt[final_ctr] += sum;
printf(\"%d %s\",final_cnt[final_ctr],final[final_ctr]);
}
ctr++; final_ctr++;
}
i++;
if(i==lines)
{
break;
}
}
free(arr);
return 0;
}
#include
#include
#include
#include
/*Structure for storing words received from each thread*/
struct return_val{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*Aggregates Frequency Sum of different threads*/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/*Function executed by each thread on separate line*/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] .
Program to sort the n names in an alphabetical order
It is a cpp program.Sort names by alphabetical order.
Users give some names as input and they will get sorted name by alphabetical order.I think that it may be helpful to others
Recommended
C++ PROGRAMThis program builds on the code below#include iostr.pdf
C++ PROGRAM
This program builds on the code below:
#include
#include
#include \"shirtType.h\"
using namespace std;
void getInput(shirtType s[],ifstream& infile, int& numShirts);
int main()
{
int numShirts;
shirtType shirts[500];
ifstream infile;
infile.open(\"h:\\\\shirtlab.txt\");
getInput(shirts,infile,numShirts);
for(int i =0; i>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
while(infile)
{
s[index].setAll(id, c, size, p, q);
index++;
infile>>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
}
numShirts=index;
}
See also the description of the class shirtType below:
#include
#include
using namespace std;
class shirtType
{
private:
int shirtID;
string color;
string size;
double price;
int quantity;
public:
void setAll(int i,string c,string s, double p, int q);
void print() const;
void newQuan();
shirtType();
shirtType(int,string,string,double,int);
};
shirtType::shirtType(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
shirtType::shirtType()
{
shirtID=0;
color=\"\";
size=\"\";
price=0.0;
quantity=0;
}
void shirtType::setAll(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
void shirtType::newQuan()
{
cout<<\"Enter the new quantity of shirts: \"<>quantity;
}
void shirtType::print() const
{
cout<<\"Shirt ID: \"<
Solution
#include
#include
struct node
{
int data;
struct node *next;
}*head;
void append(int nkkm)
{
struct node *temp,*right;
temp= (struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
right=(struct node *)head;
while(right->next != NULL)
right=right->next;
right->next =temp;
right=temp;
right->next=NULL;
}
void add( int nkkm )
{
struct node *temp;
temp=(struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
if (head== NULL)
{
head=temp;
head->next=NULL;
}
else
{
temp->next=head;
head=temp;
}
}
void addafter(int nkkm, int loc)
{
int i;
struct node *temp,*left,*right;
right=head;
for(i=1;inext;
}
temp=(struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
left->next=temp;
left=temp;
left->next=right;
return;
}
void addaShirt(int nkkm)
{
int c=0;
struct node *temp;
temp=head;
if(temp==NULL)
{
add(nkkm);
}
else
{
while(temp!=NULL)
{
if(temp->datanext;
}
if(c==0)
add(nkkm);
else if(cdata==nkkm)
{
if(temp==head)
{
head=temp->next;
free(temp);
return 1;
}
else
{
prev->next=temp->next;
free(temp);
return 1;
}
}
else
{
prev=temp;
temp= temp->next;
}
}
return 0;
}
void display(struct node *r)
{
r=head;
if(r==NULL)
{
return;
}
while(r!=NULL)
{
printf(\"%d \",r->data);
r=r->next;
}
printf(\"\ \");
}
int count()
{
struct node *n;
int c=0;
n=head;
while(n!=NULL)
{
n=n->next;
c++;
}
return c;
}
int main()
{
int i,nkkm;
struct node *n;
head=NULL;
while(1)
{
printf(\"1.Add a shirt\ \");
printf(\"2.Sell a shirt\ \");
printf(\"3.Res.
Basic C Programming Lab Practice
In this slide, we discussed the assignment of solving known programming problems.
Programming Fundamentals Arrays and Strings
Programming Fundamentals
Arrays and Strings
• Arrays
• Initializing arrays
• Multidimensional arrays
• Arrays as arguments to functions
• Strings
• String functions
Accessing Individual Components
Declaring Arrays
Arrays: Example Garbage
Multidimensional Arrays
1) Write a C program using arrays that produces the multiplication of two 2x2 matrices.
3-Dimensional Array
Write a program of your own choice that makes use of arrays of more than 2 dimensions.
2-Dimensional Arrays
Twitter sentiment analysis for cryptoassets
python code
Please email at: mamoon@quake.vc or mik279@nyu.edu for questions about how to make modifications.
primary code credit: @ZeroVariance twitter handle.
#include stdio.h#include stdlib.h#include string.h#inclu.pdf
#include
#include
#include
#include
/*Structure for storing words received from each thread*/
struct return_val{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*Aggregates Frequency Sum of different threads*/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/*Function executed by each thread on separate line*/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] = \"p\' \'file.txt\'\\0\";
char command[100];
command[0] = \'\\0\';
strcat(command,cmd_p1);
strcat(command,cmd_p2);
strcat(command,cmd_p3);
//usleep(line_number);
char cmd[100] = \" | tr [:space:] \'\\\ \' | grep -v \'^\\\\s*$\' | sort | uniq -c | sort\\0\";
strcat(command,cmd);
FILE *in;
in= popen(command, \"r\");
rewind(in);
char buff[50];
int counter = 0;
while(fgets(buff,sizeof(buff),in))
{
char c=\' \';
int i = 0;
int cnt = atoi(buff);
arr[line_number-1].count[counter] = cnt;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+6];
i++;
}
int cnnt = 0;
while(c!=\' \')
{
c = buff[cnnt];
cnnt++;
}
i=0;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+cnnt];
i++;
}
sprintf(arr[line_number-1].wordlist[counter],\"%s\",buff);
counter++;
}
arr[line_number-1].count[counter] = -1;
fclose(in);
return NULL;
}
int main(void)
{
FILE *fp;
fp = fopen(\"file.txt\",\"r\");
char a[200];
lines = 0;
void *status = NULL;
while(fgets(a,sizeof(a),fp) !=NULL)
{
lines++;
}
fclose(fp);
arr = (struct return_val*) calloc(lines, sizeof(struct return_val) * lines);
threads = malloc(sizeof(pthread_t)*lines);
int i;
for(i=0;i= 0)
{
if(arr[i].count[ctr] > 0)
{
strcpy(final[final_ctr],arr[i].wordlist[ctr]);
final_cnt[final_ctr] = arr[i].count[ctr];
arr[i].count[ctr] = 0;
int sum = aggregate(arr[i].wordlist[ctr]);
final_cnt[final_ctr] += sum;
printf(\"%d %s\",final_cnt[final_ctr],final[final_ctr]);
}
ctr++; final_ctr++;
}
i++;
if(i==lines)
{
break;
}
}
free(arr);
return 0;
}
#include
#include
#include
#include
/*Structure for storing words received from each thread*/
struct return_val{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*Aggregates Frequency Sum of different threads*/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/*Function executed by each thread on separate line*/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] .
Program to sort the n names in an alphabetical order
It is a cpp program.Sort names by alphabetical order.
Users give some names as input and they will get sorted name by alphabetical order.I think that it may be helpful to others
Using standard libraries like stdio and sdtlib.h and using stats.h a.pdf
Unlike the Federal income tax, most states do not require some form of state income tax.
True
False
Solution
Correct answer is True
Unlike Federal income tax, most states do not require some form of state income tax - This
statement is true as fililng of Fedral income tax return is mandatory for almost everybody but
some US states like Alaska, Florida, Navada, South Dakota, Texas, Washington and Wayoming
do not require filing of state income tax..
C-Program Custom Library, Header File, and Implementation FilesI .pdf
C-Program: Custom Library, Header File, and Implementation Files
I have this main.c file:
#include
#include
#include \"stats.h\"
int readGrades(char fileName[], float grades[]);
//Include this function prototype
void WelcomeToTheProgram();
int main()
{
///Declare Variables
char fileName[30];
float grades[1000]; //to store grades
int i, ArrayCount[10] = {0}; //to store histogram counts
/* Rewrite this line */
int num_grades; //for # of grades in file
///Call Intro Comment
WelcomeToTheProgram();
///Enter Filename
printf(\"\ \ \ Enter the data filename: \");
scanf(\"%s\", &fileName);
//Get # of grades in file
num_grades = readGrades(fileName, grades); //get # of grades in file
///Print Results
printf(\"There are %d grades read\ \", num_grades);
printf(\"Mean = %f\ \", mean(grades, num_grades));
printf(\"Variance = %f \ \", variance(grades, num_grades));
printf(\"Maximum = %f \ \", maximum(grades, num_grades));
printf(\"Minumum = %f \ \", minimum(grades, num_grades));
//Call histogram
histogram(grades, ArrayCount, num_grades);
//Print the histogram
printf(\"Grade Histogram\ \");
for (i = 0; i < 10; i++)
{
if (i != 9)
{
printf(\"\ %d %% - %d %%: %d \", i * 10, ((i + 1) * 10) - 1, ArrayCount[i]);
}
else
{
printf(\"\ 90 %% - 100 %%: %d\ \", ArrayCount[i]);
}
}
return 0;
}
///Reads all grades from the file to array
int readGrades(char fileName[], float grades[])
{
FILE *fpin = fopen(fileName, \"r\");
int num_scores = 0;
float value;
if (fpin == NULL)
{
printf(\"Cannot open the file\ \");
exit(0);
}
while ((fscanf(fpin, \"%f\", &value)) != EOF)
{
grades[num_scores] = value;
num_scores++;
}
return num_scores;
}
void WelcomeToTheProgram()
{
printf(\"***Welcome to the Program Which Involves a Custom Library***\");
printf(\"\ The program finds the minimum value, maximum value, mean,\");
printf(\"variance, and a histogram of the grades of data by calling\");
printf(\"implementation files min(), max(), mean(), variance(), and histogram().\ \");
}
But, the only thing that prints is
Enter the data file name: grades.txt
There are 400 grades read.
Then the program just stops, it does not print the min, max, mean, variance, or histogram. Any
ideas? Am I using the custom library wrong, the implementation files (for min, max, mean,
variance, and histogram) all check out with no warnings or errors.
Here is stats.h:
#ifndef STATS_H
#define STATS_H
float minimum(float *grades[], int Size); // NOTE: You need to complete the prototypes
float maximum(float *grades[], int Size);
float mean(float *grades[], int Size);
float variance(float *grades[], int Size);
void histogram(float *grades[], int *ArrayCount, int Size);
#endif // STATS_H
And here is the grades file:
Solution
//stats.h
#ifndef STATS_H
#define STATS_H
float minimum(float grades[], int Size); // NOTE: You need to complete the prototypes
float maximum(float grades[], int Size);
float mean(float grades[], int Size);
float variance(float grades[], int Size);
void histogram(float grades[], int ArrayCount[], int S.
PCA-2 Programming and Solving 2nd Sem.pdf
C language Basic Questions for 1st year, 2nd sem Engineering Students Specially for MAKAUT students
PCA-2 Programming and Solving 2nd Sem.docx
C language Basic Questions for 1st year, 2nd sem Engineering Students Specially for MAKAUT students
include ltstdiohgt include ltstringhgt define M.pdf
#include <stdio.h>
#include <string.h>
#define MAX_USERS 100
struct user {
char username[20];
char password[20];
char codeword[20];
};
int num_users = 0;
struct user users[MAX_USERS];
int find_user_index(char *username) {
for (int i = 0; i < num_users; i++) {
if (strcmp(users[i].username, username) == 0) {
return i;
}
}
return -1;
}
void add_user() {
if (num_users == MAX_USERS) {
printf("Error: maximum number of users reachedn");
return;
}
struct user u;
printf("Enter username: ");
scanf("%s", u.username);
printf("Enter password: ");
scanf("%s", u.password);
printf("Enter codeword: ");
scanf("%s", u.codeword);
users[num_users++] = u;
}
void edit_user() {
char username[20];
printf("Enter username to edit: ");
scanf("%s", username);
int i = find_user_index(username);
if (i == -1) {
printf("Error: user not foundn");
return;
}
struct user u = users[i];
printf("Enter new password (or press enter to keep current password): ");
scanf("%s", u.password);
printf("Enter new codeword (or press enter to keep current codeword): ");
scanf("%s", u.codeword);
users[i] = u;
}
void display_users() {
printf("%-20s %-20s %-20sn", "Username", "Password", "Codeword");
for (int i = 0; i < num_users; i++) {
printf("%-20s %-20s %-20sn", users[i].username, users[i].password, users[i].codeword);
}
}
void remove_user() {
char username[20];
printf("Enter username to remove: ");
scanf("%s", username);
int i = find_user_index(username);
if (i == -1) {
printf("Error: user not foundn");
return;
}
for (int j = i; j < num_users - 1; j++) {
users[j] = users[j + 1];
}
num_users--;
}
int main() {
int choice;
while (1) {
printf("n");
printf("1. Add usern");
printf("2. Edit usern");
printf("3. Display usersn");
printf("4. Remove usern");
printf("5. Exitn");
printf("Enter choice: ");
scanf("%d", &choice);
switch (choice) {
case 1:
add_user();
break;
case 2:
edit_user();
break;
case 3:
display_users();
break;
case 4:
remove_user();
break;
case 5:
return 0;
default:
printf("Error: invalid choicen");
}
}
}
Based on the code above, provide the necessary comments in the code..
Compare and contrast the advantages and disadvantages of arrays and .pdf
Compare and contrast the advantages and disadvantages of arrays and linked lists. Give an
example problem that is best accomplished by an array. Give an example problem that is best
accomplished with an array.
access directly with array while linked list take some time to access data. However, list can insert
easily the value into middle of list while array needs to increase the capacity and copy behind
part and fore part and input. Array\'s size is fixed while linked list has flexible size.
Stack or queue is well match with linked list because if push and pop from the end point of data
structure.
array is good with when program needs random access.
Solution
Ex: 1,2,3,4,5 to be stored in array
a[0]=1, a[1]=2 etc..
in LL we have to create space for 1 and store the address next to it as null. next when we want to
insert 2 we have to create address for 2 and store that address in pointer of 1. and so on==> 1->2-
>3->4->5.
Initiallly array size is to be definrd, but ll size can grown dynamicallyArrayLinked
ListCollections of elements with same data typeOrdered collection of data conecting by
linksArray elements are stored continuosly in memorystored in the space available and are
linkedcan have random acessbut here it will be sequential accesstakes more time for insertion
and deletion because we are storing continuosly in the memoryit takes more time here as the
value of the link needs to changed onlyIt is static memory allocation as we have to decide
beforeDynamic we have to create space for each value storedCan be 1D 2D and 3DSingular,
Doubly, circularelement is independent because we do not have connection with previous
memory locationaddress of the element is stored in the link partneed extra space for pointerneed
extra space for pointer.
Answer and describe the following five plants habit, habitat, life .pdf
Answer and describe the following five plants: habit, habitat, life span, throns, spines or prickles,
infloresences, leaves (complexity, attachment to stem, arrangement, blade shape, margin, apex,
base, venation and trichomes). Other interesting facts.
1. Isomeris arboea (Bladderpod) - Drought tolerant plant
2. Sinningia macrostachya
3. Ceropegia dichotoma
4. Lilium longiflorum
5. Peperomia verticiliata
Solution
I. Isomeris arboea
Also known as Bladderpod, Burrofat and California cleome.This plant is a fast-growing
evergreen dicot shrub from the family Capparaceae. It is a dense shrub with profuse branching
and small hairs/trichomes. It normally grows in well-draining rocky areas with good exposure to
sunlight. It is found in Southern Sierra Nevada Foothills, Tehachapi Mountain area, San Joaquin
Valley,Central Coast, South Coast, Channel Islands, Deserts, and Baja,CA
Habitat - Coastal bluffs, hills, desert washes, flats below 3,900\'
Height by Width: 3-4\' H x 4\' W
Stems - profusely branched; glabrate or puberulent (bark corky, twigs smooth)
inflorescence: The plant produces abundant inflorescences at the ends of the stem branches, each
a cluster of bright yellow flowers. Each flower has usually four petals and six whiskery
protruding stamens with curling tips holding the anthers. The fruit is an inflated capsule about 4
centimeters long and usually oval in shape. It is smooth and green when new, aging to light
brown. Fruits when dry, resemble a paper lantern that rattles (the seeds inside the dried fruit
rattle around). The flowers are dense terminal racemes. The calyx is four-cleft and the sepals are
fused in the basal half. It has four petals ½” long, 6 yellow stamens that are long and a pistil with
a short style. Sepals persistent, connate ca. 1/2 of length, green, lanceolate, 4–7 × 2.2–4 mm,
margins entire, glabrous; petals yellow, ovate-elliptic, 8–14 × 4.2–5 mm, (apex acute); stamens
yellow, 15–25 mm; anthers 2–2.5 mm; gynophore (reflexed), 10–20 mm in fruit; ovary 3–6 mm
(often aborting in bud); style 0.9–1.2 mm. Capsules (tardily dehiscent), usually inflated ,(valves
sometimes 3), 20–30 × 6–12 mm, smooth. Seeds 5–25, dark brown, obovoid, 6–7 × 5–6 mm,
smooth.
Leaves: Its leaves are made up of three equal leaflike leaflets, each a long, pointed oval 1-4 cm
long. The leaves are alternate, entire, petiolate and trifoliate with leaflets that are oblong-elliptic
with small pointed tips. petiole 1–3 cm; leaflets 3, blade oblong-elliptic, 1.5–4.5 × 0.4–1.3 cm,
margins serrate, apex acuminate to obtuse, surfaces glaucous. Racemes 1–3 cm (6–40 cm in
fruit); bracts unifoliate, obovate to spatulate, 2–15 mm. Pedicels 7–15 mm (thickened in fruit).
extra information:
II. Sinningia macrostachya
Family: Gesneriaceae
Popular names – leather leaf, Gloxinia
Sinningia macrostachya has a perennial stem base, very stiff leaves, and numerous small orange
flowers. Its fleshy stems are joined to the woody trunk. Each year, the woody part of the stems
exten.
A plant has the genotype (Aa; Bb; Cc). How many different genotype.pdf
A plant has the genotype (A/a; B/b; C/c). How many different genotypes can haploid cells
obtain after meiosis?
Solution
Answer: 8 different types.
Explanation:
Types of different genotype = 2^n,
n is the number of heterozygous alleles
In the given genotype = AaBbCc there 3 heterozygous alleles
Therefore 2^ 3 = 8 different types of gametes.
========.
A) Explain how bilateral symmetry contributed to motion and cephaliz.pdf
A) Explain how bilateral symmetry contributed to motion and cephalization (the evolution of a
head and brain)? B) Explain the common adaptations for reproduction seen in both: organisms
with a parasitic life-style and sessile organisms (those that do not move). C) Explain why those
adaptations are beneficial for each life-style.
Solution
When the animal has right and left sides morror images of each other such a symettry is called
bilateral symmetry.
Bilateral symmetry has contributed to both cephalization and locomotion in animals possessing
them. Animals wuth bilateral symmetry has locomotive organs legs on opposite sides of the body
in symmetrical position. This symmetrical arrangement of locomotive organs enables these
animalsto move thotougjly in steady direction.
Bilateral symmetry resulted in the concentration of brain ,nerves, sense organs ,eyes to the
anterior side of the body forming head..
More Related Content
Similar to #include iostream #include fstream #include shirtType.h.pdf
Using standard libraries like stdio and sdtlib.h and using stats.h a.pdf
Unlike the Federal income tax, most states do not require some form of state income tax.
True
False
Solution
Correct answer is True
Unlike Federal income tax, most states do not require some form of state income tax - This
statement is true as fililng of Fedral income tax return is mandatory for almost everybody but
some US states like Alaska, Florida, Navada, South Dakota, Texas, Washington and Wayoming
do not require filing of state income tax..
C-Program Custom Library, Header File, and Implementation FilesI .pdf
C-Program: Custom Library, Header File, and Implementation Files
I have this main.c file:
#include
#include
#include \"stats.h\"
int readGrades(char fileName[], float grades[]);
//Include this function prototype
void WelcomeToTheProgram();
int main()
{
///Declare Variables
char fileName[30];
float grades[1000]; //to store grades
int i, ArrayCount[10] = {0}; //to store histogram counts
/* Rewrite this line */
int num_grades; //for # of grades in file
///Call Intro Comment
WelcomeToTheProgram();
///Enter Filename
printf(\"\ \ \ Enter the data filename: \");
scanf(\"%s\", &fileName);
//Get # of grades in file
num_grades = readGrades(fileName, grades); //get # of grades in file
///Print Results
printf(\"There are %d grades read\ \", num_grades);
printf(\"Mean = %f\ \", mean(grades, num_grades));
printf(\"Variance = %f \ \", variance(grades, num_grades));
printf(\"Maximum = %f \ \", maximum(grades, num_grades));
printf(\"Minumum = %f \ \", minimum(grades, num_grades));
//Call histogram
histogram(grades, ArrayCount, num_grades);
//Print the histogram
printf(\"Grade Histogram\ \");
for (i = 0; i < 10; i++)
{
if (i != 9)
{
printf(\"\ %d %% - %d %%: %d \", i * 10, ((i + 1) * 10) - 1, ArrayCount[i]);
}
else
{
printf(\"\ 90 %% - 100 %%: %d\ \", ArrayCount[i]);
}
}
return 0;
}
///Reads all grades from the file to array
int readGrades(char fileName[], float grades[])
{
FILE *fpin = fopen(fileName, \"r\");
int num_scores = 0;
float value;
if (fpin == NULL)
{
printf(\"Cannot open the file\ \");
exit(0);
}
while ((fscanf(fpin, \"%f\", &value)) != EOF)
{
grades[num_scores] = value;
num_scores++;
}
return num_scores;
}
void WelcomeToTheProgram()
{
printf(\"***Welcome to the Program Which Involves a Custom Library***\");
printf(\"\ The program finds the minimum value, maximum value, mean,\");
printf(\"variance, and a histogram of the grades of data by calling\");
printf(\"implementation files min(), max(), mean(), variance(), and histogram().\ \");
}
But, the only thing that prints is
Enter the data file name: grades.txt
There are 400 grades read.
Then the program just stops, it does not print the min, max, mean, variance, or histogram. Any
ideas? Am I using the custom library wrong, the implementation files (for min, max, mean,
variance, and histogram) all check out with no warnings or errors.
Here is stats.h:
#ifndef STATS_H
#define STATS_H
float minimum(float *grades[], int Size); // NOTE: You need to complete the prototypes
float maximum(float *grades[], int Size);
float mean(float *grades[], int Size);
float variance(float *grades[], int Size);
void histogram(float *grades[], int *ArrayCount, int Size);
#endif // STATS_H
And here is the grades file:
Solution
//stats.h
#ifndef STATS_H
#define STATS_H
float minimum(float grades[], int Size); // NOTE: You need to complete the prototypes
float maximum(float grades[], int Size);
float mean(float grades[], int Size);
float variance(float grades[], int Size);
void histogram(float grades[], int ArrayCount[], int S.
PCA-2 Programming and Solving 2nd Sem.pdf
C language Basic Questions for 1st year, 2nd sem Engineering Students Specially for MAKAUT students
PCA-2 Programming and Solving 2nd Sem.docx
C language Basic Questions for 1st year, 2nd sem Engineering Students Specially for MAKAUT students
include ltstdiohgt include ltstringhgt define M.pdf
#include <stdio.h>
#include <string.h>
#define MAX_USERS 100
struct user {
char username[20];
char password[20];
char codeword[20];
};
int num_users = 0;
struct user users[MAX_USERS];
int find_user_index(char *username) {
for (int i = 0; i < num_users; i++) {
if (strcmp(users[i].username, username) == 0) {
return i;
}
}
return -1;
}
void add_user() {
if (num_users == MAX_USERS) {
printf("Error: maximum number of users reachedn");
return;
}
struct user u;
printf("Enter username: ");
scanf("%s", u.username);
printf("Enter password: ");
scanf("%s", u.password);
printf("Enter codeword: ");
scanf("%s", u.codeword);
users[num_users++] = u;
}
void edit_user() {
char username[20];
printf("Enter username to edit: ");
scanf("%s", username);
int i = find_user_index(username);
if (i == -1) {
printf("Error: user not foundn");
return;
}
struct user u = users[i];
printf("Enter new password (or press enter to keep current password): ");
scanf("%s", u.password);
printf("Enter new codeword (or press enter to keep current codeword): ");
scanf("%s", u.codeword);
users[i] = u;
}
void display_users() {
printf("%-20s %-20s %-20sn", "Username", "Password", "Codeword");
for (int i = 0; i < num_users; i++) {
printf("%-20s %-20s %-20sn", users[i].username, users[i].password, users[i].codeword);
}
}
void remove_user() {
char username[20];
printf("Enter username to remove: ");
scanf("%s", username);
int i = find_user_index(username);
if (i == -1) {
printf("Error: user not foundn");
return;
}
for (int j = i; j < num_users - 1; j++) {
users[j] = users[j + 1];
}
num_users--;
}
int main() {
int choice;
while (1) {
printf("n");
printf("1. Add usern");
printf("2. Edit usern");
printf("3. Display usersn");
printf("4. Remove usern");
printf("5. Exitn");
printf("Enter choice: ");
scanf("%d", &choice);
switch (choice) {
case 1:
add_user();
break;
case 2:
edit_user();
break;
case 3:
display_users();
break;
case 4:
remove_user();
break;
case 5:
return 0;
default:
printf("Error: invalid choicen");
}
}
}
Based on the code above, provide the necessary comments in the code..
Similar to #include iostream #include fstream #include shirtType.h.pdf (18)
Using standard libraries like stdio and sdtlib.h and using stats.h a.pdf
Using standard libraries like stdio and sdtlib.h and using stats.h a.pdf
C-Program Custom Library, Header File, and Implementation FilesI .pdf
C-Program Custom Library, Header File, and Implementation FilesI .pdf
include ltstdiohgt include ltstringhgt define M.pdf
include ltstdiohgt include ltstringhgt define M.pdf
More from arihantpatna
Compare and contrast the advantages and disadvantages of arrays and .pdf
Compare and contrast the advantages and disadvantages of arrays and linked lists. Give an
example problem that is best accomplished by an array. Give an example problem that is best
accomplished with an array.
access directly with array while linked list take some time to access data. However, list can insert
easily the value into middle of list while array needs to increase the capacity and copy behind
part and fore part and input. Array\'s size is fixed while linked list has flexible size.
Stack or queue is well match with linked list because if push and pop from the end point of data
structure.
array is good with when program needs random access.
Solution
Ex: 1,2,3,4,5 to be stored in array
a[0]=1, a[1]=2 etc..
in LL we have to create space for 1 and store the address next to it as null. next when we want to
insert 2 we have to create address for 2 and store that address in pointer of 1. and so on==> 1->2-
>3->4->5.
Initiallly array size is to be definrd, but ll size can grown dynamicallyArrayLinked
ListCollections of elements with same data typeOrdered collection of data conecting by
linksArray elements are stored continuosly in memorystored in the space available and are
linkedcan have random acessbut here it will be sequential accesstakes more time for insertion
and deletion because we are storing continuosly in the memoryit takes more time here as the
value of the link needs to changed onlyIt is static memory allocation as we have to decide
beforeDynamic we have to create space for each value storedCan be 1D 2D and 3DSingular,
Doubly, circularelement is independent because we do not have connection with previous
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Answer and describe the following five plants habit, habitat, life .pdf
Answer and describe the following five plants: habit, habitat, life span, throns, spines or prickles,
infloresences, leaves (complexity, attachment to stem, arrangement, blade shape, margin, apex,
base, venation and trichomes). Other interesting facts.
1. Isomeris arboea (Bladderpod) - Drought tolerant plant
2. Sinningia macrostachya
3. Ceropegia dichotoma
4. Lilium longiflorum
5. Peperomia verticiliata
Solution
I. Isomeris arboea
Also known as Bladderpod, Burrofat and California cleome.This plant is a fast-growing
evergreen dicot shrub from the family Capparaceae. It is a dense shrub with profuse branching
and small hairs/trichomes. It normally grows in well-draining rocky areas with good exposure to
sunlight. It is found in Southern Sierra Nevada Foothills, Tehachapi Mountain area, San Joaquin
Valley,Central Coast, South Coast, Channel Islands, Deserts, and Baja,CA
Habitat - Coastal bluffs, hills, desert washes, flats below 3,900\'
Height by Width: 3-4\' H x 4\' W
Stems - profusely branched; glabrate or puberulent (bark corky, twigs smooth)
inflorescence: The plant produces abundant inflorescences at the ends of the stem branches, each
a cluster of bright yellow flowers. Each flower has usually four petals and six whiskery
protruding stamens with curling tips holding the anthers. The fruit is an inflated capsule about 4
centimeters long and usually oval in shape. It is smooth and green when new, aging to light
brown. Fruits when dry, resemble a paper lantern that rattles (the seeds inside the dried fruit
rattle around). The flowers are dense terminal racemes. The calyx is four-cleft and the sepals are
fused in the basal half. It has four petals ½” long, 6 yellow stamens that are long and a pistil with
a short style. Sepals persistent, connate ca. 1/2 of length, green, lanceolate, 4–7 × 2.2–4 mm,
margins entire, glabrous; petals yellow, ovate-elliptic, 8–14 × 4.2–5 mm, (apex acute); stamens
yellow, 15–25 mm; anthers 2–2.5 mm; gynophore (reflexed), 10–20 mm in fruit; ovary 3–6 mm
(often aborting in bud); style 0.9–1.2 mm. Capsules (tardily dehiscent), usually inflated ,(valves
sometimes 3), 20–30 × 6–12 mm, smooth. Seeds 5–25, dark brown, obovoid, 6–7 × 5–6 mm,
smooth.
Leaves: Its leaves are made up of three equal leaflike leaflets, each a long, pointed oval 1-4 cm
long. The leaves are alternate, entire, petiolate and trifoliate with leaflets that are oblong-elliptic
with small pointed tips. petiole 1–3 cm; leaflets 3, blade oblong-elliptic, 1.5–4.5 × 0.4–1.3 cm,
margins serrate, apex acuminate to obtuse, surfaces glaucous. Racemes 1–3 cm (6–40 cm in
fruit); bracts unifoliate, obovate to spatulate, 2–15 mm. Pedicels 7–15 mm (thickened in fruit).
extra information:
II. Sinningia macrostachya
Family: Gesneriaceae
Popular names – leather leaf, Gloxinia
Sinningia macrostachya has a perennial stem base, very stiff leaves, and numerous small orange
flowers. Its fleshy stems are joined to the woody trunk. Each year, the woody part of the stems
exten.
A plant has the genotype (Aa; Bb; Cc). How many different genotype.pdf
A plant has the genotype (A/a; B/b; C/c). How many different genotypes can haploid cells
obtain after meiosis?
Solution
Answer: 8 different types.
Explanation:
Types of different genotype = 2^n,
n is the number of heterozygous alleles
In the given genotype = AaBbCc there 3 heterozygous alleles
Therefore 2^ 3 = 8 different types of gametes.
========.
A) Explain how bilateral symmetry contributed to motion and cephaliz.pdf
A) Explain how bilateral symmetry contributed to motion and cephalization (the evolution of a
head and brain)? B) Explain the common adaptations for reproduction seen in both: organisms
with a parasitic life-style and sessile organisms (those that do not move). C) Explain why those
adaptations are beneficial for each life-style.
Solution
When the animal has right and left sides morror images of each other such a symettry is called
bilateral symmetry.
Bilateral symmetry has contributed to both cephalization and locomotion in animals possessing
them. Animals wuth bilateral symmetry has locomotive organs legs on opposite sides of the body
in symmetrical position. This symmetrical arrangement of locomotive organs enables these
animalsto move thotougjly in steady direction.
Bilateral symmetry resulted in the concentration of brain ,nerves, sense organs ,eyes to the
anterior side of the body forming head..
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A. Field
B. Aggregation
C. Composite key
D. Foreign key
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D. The foreign key because it refers to an attribute in one table whose value matches the primary
key in another table. Answer A is incorrect because the field refers to the smallest unit of data
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(2, 2) and test (4) yield the same value? Add a row to test so that the last row of test contains the
entries 2, 9 Create a 1x2 matrix called test 1 whose elements are 12 and 14. Concatenate test 1
with test. Use the cat function to add another column to test. The elements of this new column
are 4, 3, and 7
Solution
(a)
%(a) creating 2 by 2 test matrix
for i=1:2
for j=1:2
test(i,j)=i^2-j;
end
end
test
test(1,2)=-5;
test(2,2)=20;
test
(b)
test(3)=-5 (Since third element matrix is -5 or actual index test(2,1) so 2+1=3, test(3)=-5).
test(5) displays error message because it contains only 4 elements.
test(2,2) and test(4) both are same because index (2,2) means (2+2)=4. so test(2,2)=test(4)
(c)
test(3,1)=2;
test(3,2)=9;
(d)
test1=[12 14];
test=[test; test1].
Write a program in Java that reads a set of doubles from a file, sto.pdf
will you give your wife your money
Solution
Yes, after marriage the couple is a family, and every individual in the family has basic needs
which can be fulfilled only by money. Even when only the husband earns, the income is meant
for the family, as he is equally responsible for each and every member of the family. However it
is better, if every individual in the family earns independently..
Which of the following statements is true about cultureA. Culture.pdf
Which is the appropriate test (Z test, single sample, independent samples t or related samples t)
for the following: A researcher is interested in determining whether a new med for appetite
suppression is effective by giving the med to one group that has been matched on weight with
another group that reveives a placebo.
Solution
As the medicine is given to two groups and we want to test whether a new med for appetite
suppression is effective or not. This can be tested only by considering the data of the two
samples and taking their difference.
So, the test has to be a two sample test. So, the possibility of a Z test or a single sample t-test is
nullified. Next, as the two sample data are matched by weight, so the data of the two samples are
no more independent. Hence, the option of independent t-test is also nullified.
So, we are only left with the related sample t-test. And this is also supported by the data. As they
are matched by weight, so they are related to each other. Hence we would go for aa \'related
sample t-test\'..
Which data set has the least sample standard deviation Data set (ii.pdf
What specific mutation is in each of the following proto-oncogenes: A)EGF receptor, B) Ras, C)
Src kinase, would cause them to become oncogenes? Explain.
Solution
A) The most common EGFR mutations are deletion of exon 19 which converts the EGFR proto
oncogene to oncogene that leads to lung cancer. B) Ras proto-oncogene encodes a protein that
receives and transfers signals required for cell division. Mutation at amino acid position 12 or 61
causes formation of oncoprotein that leads to uncontrolled cell division. Thus mutation of amino
acid 12 or 61 converts Ras proto-oncogene to oncogene. C) A truncating mutation of codon 531
of SRC proto-oncogene converts it into oncogene that leads to colorectal cancer.
Whats wrong with this Hardy Weinberg question I thought it was a.pdf
What is the price earnings Ratio for Lowe\'s and Home Depot?
HOME DEPOT Balance Sheet
Consolidated Balance Sheets - USD ($)
$ in Millions
LOWES Balance Sheet
Consolidated Balance Sheets - USD ($)
$ in Millions
Consolidated Balance Sheets - USD ($)
$ in Millions Jan. 29, 2017 Jan. 31, 2016Current Assets:Cash and Cash Equivalents$ 2,538$
2,216Receivables, net2,0291,890Merchandise Inventories12,54911,809Other Current
Assets608569Total Current Assets17,72416,484Property and Equipment, at
cost40,42639,266Less Accumulated Depreciation and Amortization18,51217,075Net Property
and Equipment21,91422,191Goodwill2,0932,102Other Assets1,2351,196Total
Assets42,96641,973Current Liabilities:Short-Term Debt710350Accounts
Payable7,0006,565Accrued Salaries and Related Expenses1,4841,515Sales Taxes
Payable508476Deferred Revenue1,6691,566Income Taxes Payable2534Current Installments of
Long-Term Debt54277Other Accrued Expenses2,1951,941Total Current
Liabilities14,13312,524Long-Term Debt, excluding current installments22,34920,789Other
Long-Term Liabilities1,8551,965Deferred Income Taxes296379Total
Liabilities38,63335,657STOCKHOLDERS\' EQUITYCommon Stock, par value $0.05;
authorized: 10 billion shares; issued: 1.776 billion shares at January 29, 2017 and 1.772 billion
shares at January 31, 2016; outstanding: 1.203 billion shares at January 29, 2017 and 1.252
billion shares at January 31, 20168888Paid-In Capital9,7879,347Retained
Earnings35,51930,973Accumulated Other Comprehensive Loss(867)(898)Treasury Stock, at
cost, 573 million shares at January 29, 2017 and 520 million shares at January 31,
2016(40,194)(33,194)Total Stockholders’ Equity4,3336,316Total Liabilities and Stockholders\'
Equity$ 42,966$ 41,973
Solution
Price at 29 Jan 2017 = 40194 / 573 = $70.15 per share (from Treasury stock data)
Price at 31 Jan 2016 = 33194 / 520 = $63.83 per share
EPS for 2017 = (RE $35519 - RE $30973)m = 4546m / outstanding shares 1776m = $2.56
pershare
Price Earnings ratio for 2017 = Price on 29 Jan 17 / EPS for 2017 = $70.15 / $2.56 = 27.40 times.
6. List and define the underlying conditions, or assumptions, which .pdf
6. List and define the underlying conditions, or assumptions, which have had an impact on the
development of GAAP.
Solution
ANSWER:
The underlying conditions/ assumptions, which have had an impact on the development of
GAAP are as follows:
--Business entity: The company for which the financial statements are prepared is distinct and
separate from the owners.
--Going concern or continuity : The assumption is made that the organisation will remain in
business for an indefinite time frame.
--Monetary unit : Accountants requires certain standard of measure to bring financial
transactions together in an appropriate way.
--Realization: Revenue must be recognized when the earning process is virtually complete and
the exchange value need to be objectively determined
--Matching: This concept states to recognize the costs that are related with the recognized
revenue
--Full disclosure: Accounting reports must disclose all facts that may influence the judgment of
an informed reader
--Materiality: The approach involves the relative size and importance of an item to a business.
--Verifiability: The accountant need to adhere as closely as possible to verifiable data.
--Industry practices: Few business use accounting reports that do not conform to the general
theory that underlies accounting
--Transaction approach : The accountant records only events that affect the business financial
position and also can be reasonably determined in monetary terms.
--Accrual basis: Expenses need to recognized when incurred (matching concept), and revenue
are to be recognized when it is realized (realization concept), and
--Consistency: The business must give the similar treatment to comparable transactions from
period to period
--Conservatism : The measurement with the least favorable effect on financial position and net
income in the current period need to be selected.
--Historical cost : Of the numerous values that could be used, the objective and determinable
value should be selected
-- Time period With this assumption, inaccuracies of accounting for entity short of its complete
life span are accepted..
What does the LP tableau below indicate (The xs are decision varia.pdf
Wha is the difference between the large intestine and colon?
Wha is the difference between the large intestine and colon?
Solution
Answer:
The large intestine starts in the pelvis at the right iliac region located immediately under the
waist on the right-hand side. It has 5 parts that are -cecum, appendix, colon, rectum and anus.
Overall function of large intestine is to form feces.
Colon makes up most part of large intestine. It helps in water and salt absorption..
Wha is the difference between the large intestine and colon .pdf
Using the six general trends of the general environment, give a hypothetical example of how
each trend could affect a restaurant business.
Solution
There are six general trends of the general environment--
1) Demographic Segment--It consists of people from different ethnic groups like Indians,
Europeans, Asians etc and its affect on a business. In case of a restaurant business opening of a
Indian restaurant in China will attract lower customers
2)Socio Cultural Segment--It consists of diverse set of people in the workplace who helps in the
company\'s success in the Global Market. E.g. a business employing women, not engaging into
racial discrimination etc. will attract more employees
3)Economic Segment--It depends on the nation\'s economic condition in the World Market. In
case of a restaurant business, an economic slowdown or recession will have a negative impact on
the business.
4)Legal Segment--This segment is dependent on an economy\'s governance laws, policies,
taxation laws and regulations for its employees. In case of restaurant business, a restaurant set up
in a country with heavy taxes and strict regulation, operations can be difficult and prove to be of
losses.
5)Technological Segment--It relates to technological advancements and updation of newer
technology at pace with the competitors. In case of restaurant business, restaurants adopting
newer technologies will be able to produce more and earn more
6)Global Segment--This segment expands the organisation globally so that it can gain a
competitive advantage over its competitors when they have a global presence in the industry. In
case of a restaurant business, going global can prove to be profitable if given chances..
Was airpower decisive in winning World War II Defend your answer..pdf
Using \" for \" loop to calculate compound interest for 20 years. The starting principal is 5000
and the rate is 0.04. Show amount on deposit for each year.
Solution
#include
#include
#include
int main()
{
double principal=5000.0;
double rate=0.04;
int time=20;
double amt=principal;
for(int i=1;i<=20;i++)
{
printf(\"\ Deposit for year %d : %lf\",i,amt);
amt=amt+((amt*rate*1.0)/100);
}
printf(\"compound interest for 20 yrs : %lf\",(amt*pow(1.0+(rate/100),20)));
return 0;
}.
Summarize the workflow of the study of microbial diversity. Draw a d.pdf
Solve the following:
a. How many permutations of all 26 letters of the alphabet are there that contain at least one of
the words: CAR, CART, SCARF, SCARE, SCAR?
b. How many permutations of all 26 letters of the alphabet are there that contain at least one of
the words: MAZE, SWORD, PLANT, CARTS?
Solution
This computer science but:
a) The answer should be Factorial(24) i.e . factorial of 24. It can be solved by formulating a
simple recurrence relation.
b)(26C4+26C5+26C5+26C5)/26!.
It can also be seen as a recurrence relation and solved..
Shifting allocations most often arise as a result of which of the fo.pdf
Salaries of 40 college graduates who took a statistics course in college have a mean, x , of $
68,200. Assuming a standard deviation, , of $18,561, construct a 95% confidence interval for
estimating the population mean .
$___ < < $___
Round to the nearest integer as needed.
Solution
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=68200
Standard deviation( sd )=18561
Sample Size(n)=40
Confidence Interval = [ 68200 ± Z a/2 ( 18561/ Sqrt ( 40) ) ]
= [ 68200 - 1.96 * (2934.752) , 68200 + 1.96 * (2934.752) ]
= [ 62447.887,73952.113 ]
= [ 62448 ,73953 ].
16.) How would you determine if Cellulous is hydrophobic or hydrophi.pdf
16.) How would you determine if Cellulous is hydrophobic or hydrophilic?
Solution
Hydrophilic substance interact with water and willabsorb it easly . Hydrophobicmeans it seperate
the water . Cellulise is a plymer chain of monosachhride . It is hydrophilic and atrract water
molecules due to its functional group . But it is aplymer chai water does not have ability to
surround compltely each monomerof cellulose fiber .Polymer is less soluble than monomwes
because of entropy . cellulose have hydrogen bond but not as well as water and plymer have low
entropy due to their confuguration and constrained. the attract water on the surface if the
molecules are smaller then water . cellulose is a big polymer so the particals are to large too
dissolved..
Please Explain CompletelyWhat defines the beginning of the hepatic.pdf
Pattern of Stars
Listing 6.4 in the text book demonstrates the use of a nested for loop by printing a pattern of
asterisks (stars), starting from the least number of stars per row (1) to the most number of stars
per row (10). You will modify this program so that it does the following:
Asks the user how many rows of stars they want.
Prints the stars starting with the least stars per row and ending with the most stars per row, but
the stars must be lined up along the right margin instead of the left. (In other words, print a
version of the pattern that has been flipped horizontally i.e. along a vertical axis).
You will need:
An input statement to prompt the user for the number of rows.
An assignment statement to store the number of rows.
A nested for loop as shown in Listing 6.4 (pg 263). You will need to modify the conditions in
some of your for loops to accomplish this task.
This time, you will need two for loops nested within the outermost loop.
The outer loop will control the number of rows, as in the example.
One of the inner loops will control the number of stars per row, again as in the example.
You will need to create another for loop that controls the number of spaces per row, to achieve
the intended pattern. The spaces need to be printed before each row of stars is printed.
Note: The two inner loops do not have to be nested. See the code shell for the intended design.
Sample output:
Enter the number of rows: 8
*
**
***
****
*****
******
*******
********
Code Shell
//*********************************************************
// StarPattern.java
//
// This is a modified version of the Stars.java program
// from Listing 6.4 in the text.
//*********************************************************
//import Scanner class here
public class StarPattern1
{
//------------------------------------------------------
// Prints a triangle shape using asterisk (star)
// characters.
//------------------------------------------------------
public static void main(String[] args)
{
//Create Scanner variable
final int MAX_ROWS = 10; // Replace this constant
// with a variable to store
// the number of rows
//Ask for number of rows
//Store in variable
for (int row = 1; /*adjust condition*/; row++)
{
/*insert a for loop here. This one should
control the number of spaces before each star*/
for (int star = 1; star <= row; star++)
System.out.print(\"*\");
System.out.println();
}
}
}
Solution
//import Scanner class here
import java.util.Scanner;
public class StarPattern1{
public static void main(String[] args)
{
Scanner input = new Scanner( System.in );
//Create Scanner variable
final int MAX_ROWS;
System.out.println(\"Enter the number of rows: \");
MAX_ROWS = input.nextInt();
for (int row = 1; row=row;j--){
System.out.print(\" \"); /*insert a for loop here. This one should
control the number of spaces before each star*/
}
for (int star = 1; star <= row; star++){
System.out.print(\"*\");
}
System.out.println();
}
}
}.
Plant Diversity II – Seed Plants1. Explain how the rise in pro.pdf
ooo T-Mobile LTE 5:40 PM courses.apexlearning.com s of Polynomial Functions Which point
is the best approximation of the relative maximum of the polynomial function graphed below? 30
20 2 10 O A (-2.3, 17.9) O B. (-1.4, 16.9) O c. (-1.1, 14.8) D. (-2.1, 17.2) IL
Solution
in the x axis the value is between -1 and -2 , and the y-axis value is closer to 20 then the answer
is the letter
B = (-1.4, 16.9).
17) Fill out the following table Structure Pili or Fimbriae Flagella.pdf
17) Fill out the following table: Structure Pili or Fimbriae Flagella Capsule PHB inclusion
Sulfur granules Magnetosomes Endo LPS layer Cell wall Function Location
Solution
structure
Function
Location
Pili or fimbrae
Required for bacterial conjugation
On the surface, embedded in cell membrane
Flagella
Bacterial cell motility
On the surface, embedded in cell membrane; numbers vary from 1 to multiple
Capsule
Outside the cell envelope
Protection from antibiotics, mechanical stress and desiccation
PHB inclusion
Present in the bacterial cytoplasm
Acts as storage granules in bacteria
Sulfur granules
Inclusion bodies in cytoplasm
May aid in metabolism- bacteria that use Hydrogen sulfide as electron source for respiration
Magnetosomes
Membranous organelles that are present in magnetotactic bacteria; containing magnetic crystals
Help the bacteria orient in geomagnetic field to search proper environment
Endospores
Generally present in the cytoplasm
Helps the bacterial cells propagate asexually in extreme conditions
LPS layer
Part of cell wall in Gram negative bacterial cells
Protection of the cells, transport of materials and also imparts pathogenicity to the bacterial cells
Cell wall
A part of the cell envelope, surrounding cell membrane
Protection of the cells, transport of materials and maintenance of osmotic pressure.
structure
Function
Location
Pili or fimbrae
Required for bacterial conjugation
On the surface, embedded in cell membrane
Flagella
Bacterial cell motility
On the surface, embedded in cell membrane; numbers vary from 1 to multiple
Capsule
Outside the cell envelope
Protection from antibiotics, mechanical stress and desiccation
PHB inclusion
Present in the bacterial cytoplasm
Acts as storage granules in bacteria
Sulfur granules
Inclusion bodies in cytoplasm
May aid in metabolism- bacteria that use Hydrogen sulfide as electron source for respiration
Magnetosomes
Membranous organelles that are present in magnetotactic bacteria; containing magnetic crystals
Help the bacteria orient in geomagnetic field to search proper environment
Endospores
Generally present in the cytoplasm
Helps the bacterial cells propagate asexually in extreme conditions
LPS layer
Part of cell wall in Gram negative bacterial cells
Protection of the cells, transport of materials and also imparts pathogenicity to the bacterial cells
Cell wall
A part of the cell envelope, surrounding cell membrane
Protection of the cells, transport of materials and maintenance of osmotic pressure..
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#include iostream #include fstream #include shirtType.h.pdf
- 1. #include #include #include "shirtType.h" using namespace std; void getInput(shirtType s[],ifstream& infile, int& numShirts); int main() { int numShirts; shirtType shirts[500]; ifstream infile; infile.open("h:shirtlab.txt"); getInput(shirts,infile,numShirts); for(int i =0; i>id; infile.ignore(100,' '); getline(infile,c); infile>>size>>p>>q; while(infile) { s[index].setAll(id, c, size, p, q); index++; infile>>id; infile.ignore(100,' '); getline(infile,c); infile>>size>>p>>q; } numShirts=index; } See also the description of the class shirtType below: #include #include using namespace std; class shirtType { private:
- 2. int shirtID; string color; string size; double price; int quantity; public: void setAll(int i,string c,string s, double p, int q); void print() const; void newQuan(); shirtType(); shirtType(int,string,string,double,int); }; shirtType::shirtType(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } shirtType::shirtType() { shirtID=0; color="";
- 3. size=""; price=0.0; quantity=0; } void shirtType::setAll(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } void shirtType::newQuan() { cout<<"Enter the new quantity of shirts: "<>quantity; } void shirtType::print() const { cout<<"Shirt ID: "< Solution //shirtType.h #include #include using namespace std;
- 4. class shirtType { private: int shirtID; string color; string size; double price; int quantity; public: void setAll(int i,string c,string s, double p, int q); void print() const; void newQuan(); int changeQuan(); int getShirtId(int); shirtType(); int search(int); shirtType(int,string,string,double,int); }; int shirtType::search(int sellId) { if(shirtID == sellId) return 1; else return 0; } int shirtType::getShirtId(int i) { return shirtID; } shirtType::shirtType(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i;
- 5. else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } shirtType::shirtType() { shirtID=0; color=""; size=""; price=0.0; quantity=0; } void shirtType::setAll(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0)
- 6. quantity=q; else quantity = 0; } void shirtType::newQuan() { int q; cout<<"Enter the new quantity of shirts: "<>q; quantity += q; } int shirtType::changeQuan() { if(quantity > 0 ) { quantity -= 1; return 1; } { return 2; } } void shirtType::print() const { cout<<"Shirt ID: "< #include #include #include "shirtType.h" #include using namespace std; void getInput(shirtType s[],ifstream& infile, int& numShirts); void addShirt(shirtType s[], int& numShirts); void sellShirt(shirtType s[], int& numShirts); void addstockShirt(shirtType s[], int& numShirts); int main() { int numShirts;
- 7. char ch; shirtType shirts[500]; ifstream infile; infile.open("shirtlab.txt"); getInput(shirts,infile,numShirts); // cout<>ch; do{ switch(ch) { case 'a': addShirt(shirts,numShirts); break; case 's': sellShirt(shirts,numShirts); break; case 'r': addstockShirt(shirts,numShirts); break; case 'p': for(int i =0; i>ch; }while(ch!='q'); infile.close(); return 0; } void getInput(shirtType s[], ifstream& infile, int& numShirts) { int id, q,index=0; string c,size; double p; infile>>id; infile.ignore(100,' '); getline(infile,c); infile>>size>>p>>q; while(infile) {
- 8. s[index].setAll(id, c, size, p, q); index++; infile>>id; infile.ignore(100,' '); getline(infile,c); infile>>size>>p>>q; } numShirts=index; } void addShirt(shirtType s[], int& numShirts) { int id, q,index=numShirts; // cout<>id; cout<<" Shirt Color:"; cin>>c; // getline(cin,c); cout<<" Shirt size price and quantity:"; cin>>size>>p>>q; s[index].setAll(id, c, size, p, q); index++; numShirts=index; // cout<>sellId; for(i =0; i>stockId; for(i =0; i