increase by 3 per year thereafter If Tolo has 21 million .pdf
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increase by 3% per year thereafter. If Tolo has 2.1 million shares outstanding and an equity cost of capital of 11.1%, what is its price per share today? The stock price will be $ (Round to the nearest cent.).
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Insert a function in cell D11 to display the ltem name based.pdf
Insert a function in cell D11 to display the ltem name based on the provided inventory lookup
information in the range B24:D20. Be sure to use the appropriate absolute cell referencing and the
function returns an exact match. Copy the function from cell D11 down through D19 to complete
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Copy the formula down the column. 10 Use Quick Analysis to insert a function in cell 120 that
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include ltstdiohgt include ltstdlibhgt include .pdf
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void FillBingoCard(int bingocard[5][5]);
void PrintCard(int bingocard[5][5], int called[3], char yn[1]);
int PickNumber(int called[3], int callednumber[75], int counter);
int drawnnumber(int bingocard[5][5], int called[3]);
void completedrowcolumn(int bingocard[5][5], int win);
int main(void)
{
//array of bingo numbers
int row = 5;
int col = 5;
int bingocard[row][col];
int called[3];
int counter = 0;
int callednumber[75];
char yn[1];
int win = 0;
//fill bingocard
FillBingoCard(bingocard);
//print bingocard
PrintCard(bingocard, called, yn);
//called
called[3] = PickNumber(called, callednumber, counter);
for(int i = 0; i < 75; i++)
{
callednumber[i] = 0;
}
callednumber[counter] = called[1];
printf("nThe called number is %c%dn", called[0], called[1]);
//ask player if they have chosen number on their bingo card
printf("Do you have it? (Y/N) ");
scanf(" %c", &yn[0]);
//verify
while(counter < 75)
{
int verify;
if(yn[0] == 'Y')
{
int verify = drawnnumber(bingocard, called);
if(verify == 1)
{
counter++;
PrintCard(bingocard, called, yn);
called[3] = PickNumber(called, callednumber, counter);
callednumber[counter] = called[1];
completedrowcolumn(bingocard, win);
if(win == 0)
{
printf("nThe next number is %c%dn", called[0], called[1]);
printf("Do you have it? (Y/N) ");
scanf(" %c", &yn[0]);
}
}
else
{
printf("nThat value is not on your BINGO card - are you trying to cheat?");
counter = 76;
}
}
else
{
counter++;
PrintCard(bingocard, called, yn);
called[3] = PickNumber(called, callednumber, counter);
callednumber[counter] = called[1];
printf("nThe next number is %c%dn", called[0], called[1]);
printf("Do you have it? (Y/N) ");
scanf(" %c", &yn[0]);
}
}
if(counter == 75)
{
printf("There are no other numbers to call left.");
}
return (0);
}
void FillBingoCard(int bingocard[5][5])
{
srand(time(0));
int row, col;
int counter = 0;
for(col = 0; col < 5; col++)
{
for(row = 0; row < 5; row++)
{
if(col == 0)
{
bingocard[row][0] = (rand() % 15) + 1;
}
if(col == 1)
{
bingocard[row][1] = (rand() % 15) + 16;
}
if(col == 2)
{
bingocard[row][2] = (rand() % 15) + 31;
}
if(col == 3)
{
bingocard[row][3] = (rand() % 15) + 46;
}
if(col == 4)
{
bingocard[row][4] = (rand() % 15) + 61;
}
for(int counter = 0; counter == 0; counter = counter)
{
int check = 5;
for(int i = 0; i < 5; i++)
{
if(bingocard[row][col] == bingocard[i][col])
{
if(col == 0)
{
bingocard[row][0] = (rand() % 15) + 1;
}
if(col == 1)
{
bingocard[row][1] = (rand() % 15) + 16;
}
if(col == 2)
{
bingocard[row][2] = (rand() % 15) + 31;
}
if(col == 3)
{
bingocard[row][3] = (rand() % 15) + 46;
}
if(col == 4)
{
bingocard[row][4] = (rand() % 15) + 61;
}
}
}
for(int j = 0; j < 5; j++)
{
if(bingocard[row][col] != bingocard[j][col])
{
check--;
}
}
if(check == 1)
{
counter++;
}
}
}
}
bingocard[2][2] = 0;
}
void PrintCard(int bingocard[5][5], int called[3], char yn[1])
{
int row, col;
printf("%5c%8c%8c%8c%8cn-----------------------------------------n", 66, 73, 78, 71, 79.
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varios componentes clave de los equipos agrcolas de la empresa, incluidos los motores. Anna ha
decidido que FFE debera considerar la compra de un fabricante de motores para permitir que FFE
integre mejor su cadena de suministro y obtenga ms control sobre las caractersticas del motor.
Despus de investigar varias posibles empresas, Anna siente que la compra de Mantz Engines Inc
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empresa fabrica motores agrcolas que se utilizan principalmente en equipos agrcolas. Mantz ha
experimentado un rpido crecimiento gracias a una tecnologa patentada que aumenta la eficiencia
de combustible de sus motores con muy poco sacrificio en el rendimiento. La empresa es
propiedad a partes iguales de Sydney y Claire Mantz. El acuerdo original entre los hermanos
otorgaba a cada uno 200.000 acciones. Los Mantz no han emitido acciones adicionales ni han
vendido ninguna de sus acciones desde que fundaron la empresa. Anna le ha pedido a Aiden que
determine un valor por accin de las acciones de Mantz. Aiden ha recopilado la siguiente
informacin sobre las empresas de fabricacin de motores que cotizan en bolsa. EPS DPS PRECIO
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castigo contable el ao pasado. Sin la cancelacin, el EPS de la empresa habra sido de $2,25. El ao
pasado, Mantz Engines Inc (MEI) obtuvo un EPS de $4,00 y pag un dividendo a Sydney y Claire
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Anna le dice a Aiden que se debe usar una tasa de rendimiento requerida del 14% para evaluar el
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proveedores externos para varios componentes clave de los equipos agrcolas de la empresa,
incluidos los motores. Anna ha decidido que FFE debera considerar la compra de un fabricante de
motores para permitir que FFE integre mejor su cadena de suministro y obtenga ms control sobre
las caractersticas del motor. Despus de investigar varias posibles empresas, Anna siente que la
compra de Mantz Engines Inc (MEI) es una posibilidad. Le ha pedido a Aiden Carney que analice
el valor de Mantz. Mantz Engines, Inc., es una empresa privada fundada hace siete aos por
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Insert a function in cell D11 to display the ltem name based on the provided inventory lookup
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function returns an exact match. Copy the function from cell D11 down through D19 to complete
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lookup information. Be sure to use appropriate absolute cell referencing and the function returns
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include ltstdiohgt include ltstdlibhgt include .pdf
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void FillBingoCard(int bingocard[5][5]);
void PrintCard(int bingocard[5][5], int called[3], char yn[1]);
int PickNumber(int called[3], int callednumber[75], int counter);
int drawnnumber(int bingocard[5][5], int called[3]);
void completedrowcolumn(int bingocard[5][5], int win);
int main(void)
{
//array of bingo numbers
int row = 5;
int col = 5;
int bingocard[row][col];
int called[3];
int counter = 0;
int callednumber[75];
char yn[1];
int win = 0;
//fill bingocard
FillBingoCard(bingocard);
//print bingocard
PrintCard(bingocard, called, yn);
//called
called[3] = PickNumber(called, callednumber, counter);
for(int i = 0; i < 75; i++)
{
callednumber[i] = 0;
}
callednumber[counter] = called[1];
printf("nThe called number is %c%dn", called[0], called[1]);
//ask player if they have chosen number on their bingo card
printf("Do you have it? (Y/N) ");
scanf(" %c", &yn[0]);
//verify
while(counter < 75)
{
int verify;
if(yn[0] == 'Y')
{
int verify = drawnnumber(bingocard, called);
if(verify == 1)
{
counter++;
PrintCard(bingocard, called, yn);
called[3] = PickNumber(called, callednumber, counter);
callednumber[counter] = called[1];
completedrowcolumn(bingocard, win);
if(win == 0)
{
printf("nThe next number is %c%dn", called[0], called[1]);
printf("Do you have it? (Y/N) ");
scanf(" %c", &yn[0]);
}
}
else
{
printf("nThat value is not on your BINGO card - are you trying to cheat?");
counter = 76;
}
}
else
{
counter++;
PrintCard(bingocard, called, yn);
called[3] = PickNumber(called, callednumber, counter);
callednumber[counter] = called[1];
printf("nThe next number is %c%dn", called[0], called[1]);
printf("Do you have it? (Y/N) ");
scanf(" %c", &yn[0]);
}
}
if(counter == 75)
{
printf("There are no other numbers to call left.");
}
return (0);
}
void FillBingoCard(int bingocard[5][5])
{
srand(time(0));
int row, col;
int counter = 0;
for(col = 0; col < 5; col++)
{
for(row = 0; row < 5; row++)
{
if(col == 0)
{
bingocard[row][0] = (rand() % 15) + 1;
}
if(col == 1)
{
bingocard[row][1] = (rand() % 15) + 16;
}
if(col == 2)
{
bingocard[row][2] = (rand() % 15) + 31;
}
if(col == 3)
{
bingocard[row][3] = (rand() % 15) + 46;
}
if(col == 4)
{
bingocard[row][4] = (rand() % 15) + 61;
}
for(int counter = 0; counter == 0; counter = counter)
{
int check = 5;
for(int i = 0; i < 5; i++)
{
if(bingocard[row][col] == bingocard[i][col])
{
if(col == 0)
{
bingocard[row][0] = (rand() % 15) + 1;
}
if(col == 1)
{
bingocard[row][1] = (rand() % 15) + 16;
}
if(col == 2)
{
bingocard[row][2] = (rand() % 15) + 31;
}
if(col == 3)
{
bingocard[row][3] = (rand() % 15) + 46;
}
if(col == 4)
{
bingocard[row][4] = (rand() % 15) + 61;
}
}
}
for(int j = 0; j < 5; j++)
{
if(bingocard[row][col] != bingocard[j][col])
{
check--;
}
}
if(check == 1)
{
counter++;
}
}
}
}
bingocard[2][2] = 0;
}
void PrintCard(int bingocard[5][5], int called[3], char yn[1])
{
int row, col;
printf("%5c%8c%8c%8c%8cn-----------------------------------------n", 66, 73, 78, 71, 79.
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Informacin general Anna ha estado hablando con los directores de la empresa sobre el futuro de
Frentheway Farm Equipment (FFE). La empresa ha estado utilizando proveedores externos para
varios componentes clave de los equipos agrcolas de la empresa, incluidos los motores. Anna ha
decidido que FFE debera considerar la compra de un fabricante de motores para permitir que FFE
integre mejor su cadena de suministro y obtenga ms control sobre las caractersticas del motor.
Despus de investigar varias posibles empresas, Anna siente que la compra de Mantz Engines Inc
(MEI) es una posibilidad. Le ha pedido a Aiden Carney que analice el valor de Mantz. Mantz
Engines, Inc., es una empresa privada fundada hace siete aos por Sydney y Claire Mantz. La
empresa fabrica motores agrcolas que se utilizan principalmente en equipos agrcolas. Mantz ha
experimentado un rpido crecimiento gracias a una tecnologa patentada que aumenta la eficiencia
de combustible de sus motores con muy poco sacrificio en el rendimiento. La empresa es
propiedad a partes iguales de Sydney y Claire Mantz. El acuerdo original entre los hermanos
otorgaba a cada uno 200.000 acciones. Los Mantz no han emitido acciones adicionales ni han
vendido ninguna de sus acciones desde que fundaron la empresa. Anna le ha pedido a Aiden que
determine un valor por accin de las acciones de Mantz. Aiden ha recopilado la siguiente
informacin sobre las empresas de fabricacin de motores que cotizan en bolsa. EPS DPS PRECIO
DE LAS ACCIONES ROE ROA Jensen Motors Corp. $3,50 $2,00 $37,00 12,00% 9,00% Blue
Cow Farm Engines Inc. 8,00 4,00 90,00 15,00% 11,00% Horrocks Farm Equipment (0,50) 1,00
30,00 16,00% 13,00% Ganancias negativas por accin de Horrocks UPA) fue el resultado de un
castigo contable el ao pasado. Sin la cancelacin, el EPS de la empresa habra sido de $2,25. El ao
pasado, Mantz Engines Inc (MEI) obtuvo un EPS de $4,00 y pag un dividendo a Sydney y Claire
Mantz de $300 000 cada una. La empresa tambin tuvo un rendimiento sobre el capital del 16%.
Anna le dice a Aiden que se debe usar una tasa de rendimiento requerida del 14% para evaluar el
valor de Mantz. Informacin general Anna ha estado hablando con los directores de la empresa
sobre el futuro de Frentheway Farm Equipment (FFE). La empresa ha estado utilizando
proveedores externos para varios componentes clave de los equipos agrcolas de la empresa,
incluidos los motores. Anna ha decidido que FFE debera considerar la compra de un fabricante de
motores para permitir que FFE integre mejor su cadena de suministro y obtenga ms control sobre
las caractersticas del motor. Despus de investigar varias posibles empresas, Anna siente que la
compra de Mantz Engines Inc (MEI) es una posibilidad. Le ha pedido a Aiden Carney que analice
el valor de Mantz. Mantz Engines, Inc., es una empresa privada fundada hace siete aos por
Sydney y Claire Mantz. La empresa fabrica motores agrcolas que se utilizan principalmente en
equipos agrcolas. Mantz ha experimentado un rpido .
inflation in india interest rates in india and compare it wi.pdf
inflation in india
interest rates in india
and compare it with US federal reserve bank
what do you think US federal reserve bank is going to decide and discuss in next meeting
can you make it detail about US interest rates
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territories and resources. Globally, indigenous peoples are disproportionately affected by
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community. 2. Discuss the tension between First World and First Peoples' environmental
ideologies. Use examples from your chosen situation to demonstrate the First World view that
commodifies land versus the First Peoples view that we are the caretakers of the land. 3. Discuss
solutions to balance the needs of industrialized nations, indigenous belief systems, and
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Group of answer choices
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First prototypes are built, and then use scenarios are discussed
Use Scenarios are commonly used to begin the process of user centered design
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include ltinitializer_listgt include ltiostreamgt .pdf
#include <initializer_list>
#include <iostream>
#include <iterator>
#include <utility>
/************************************************************/
template<typename T>
struct ListNode
{
ListNode () : data ()
{
}
ListNode (const T& v) : data (v)
{
}
ListNode (const T& v, ListNode* n, ListNode* p) : data (v), next (n), prev (p)
{
}
// unhooks the range [begin,end] from a linked list
// NOTE: will lose reference to begin and end if you
// are not keeping track of it!
//
// - does not create any new nodes
// - does not destroy any existing nodes
// - begin is start of range to remove
// - end is inclusive end of range to remove
//
// [5]
static void
unhook_range (ListNode* begin, ListNode* end)
{
// TODO
}
// inserts the range [first,last] before this
// NOTE: does not create any new nodes, does not destroy any existing nodes
//
// [5]
void
hook_range (ListNode* first, ListNode* last)
{
// TODO
}
// insert first before this
void
hook (ListNode* first)
{
hook_range (first, first);
}
// unhooks current node from linked list
void
unhook ()
{
ListNode::unhook_range (this, this);
}
T data;
ListNode* next{nullptr};
ListNode* prev{nullptr};
};
Please complete the implementation of the missing part marked TODO.
// node insertion/removal [5] static Node:: unhook_range (Node* first, Node* last) [3] When first ==
last (single element) [2] When first != last [5] Node: :hook_range (Node* first, Node* last) [3] When
first != last [2] When first == last (single element).
In your opinion What would drive your decisions If you wer.pdf
In your opinion?
What would drive your decisions?
If you were deciding to invest in a company how would you use their financial statements in your
decision making process? What specific information would you review? What would you be trying
to determine?
What criteria would you use to decide whether or not to make the investment?
In addition to the financial statements what other information would you need to make your
decision or what other information would be helpful?
If you were ranking the financial statement in importance how would you rank them and why?.
Income statement for 2016 Sales are expected to increase by .pdf
Income statement for 2016
Sales are expected to increase by 10% in 2017, while costs including depreciation are expected to
reach 60% of sales. Interest expenses and the average tax rate are expected to stay constant
(in $) 2016
Sales 360,000
Costs (incl. depr.) 252,000
EBIT 108,000
Interest 25,000
Taxes (28%) 23,240
Net income 59,760.
Indicate in the figure below the location of the Stratosphe.pdf
Indicate in the figure below the location of the: Stratosphere, Tropopause, Troposphere, Boundary
layer, Surface layer
Please use locations that are appropriate for this sounding, do not use values from the USSA. You
only need to do this for the daytime sounding. You can provide the answers to this question either
in a table or by (somehow) importing the image in your document and annotating it.
Geopotential height asl (km) Altitude asl (km).
Independent ttest Dependentpaired ttest Oneway ANOVA Two.pdf
Independent t-test
Dependent/paired t-test
One-way ANOVA
Two-way ANOVA
Repeated Measures ANOVA
ANCOVA
A researcher was interested in the effects of journal therapy on depression. Sixty individuals
diagnosed with severe depression were randomly assigned to three different treatment groups (20
individuals in each group). The first group received journal therapy (writing about their feelings for
an hour each day for three weeks) and regular counseling, the second group received only journal
therapy, and the third group received only regular counseling. All sixty individuals were given a
depression scale before (baseline) and after treatment. Assume the assumptions of the parametric
method were satisfied. Do depression scores after the intervention differ among the three groups,
controlling for or removing the effect of the baseline depression score?.
Independent Assortment Two genes Dumb d and Anxious a .pdf
Independent Assortment Two genes, Dumb (d) and Anxious (a), are recessive genes routinely
found in offspring. Previous research has determined that they reside on different chromosomes.
F1 male and female are mated and their F2 offspring are scored for dumbness and anxiety.
Complete the Punnett square for the F2. Show the Phenotype, Genotype, and expected ratio for
the F2 below or to the side of the Punnett square. a. What % of the F2 offspring would you expect
that are dumb but not anxious? b. What % of the F2 offspring would you expect that are anxious
but not dumb? c. What % of the F2 offspring would you expect that are dumb and anxious? d. If
your goal is to not have offspring that are dumb or anxious, what % of the F 2 could be mated in
the F3 ?.
incubation limes that rimake up the middle 97 Cick the icon.pdf
incubation limes that rimake up the middle 97% Cick the icon to view a table of areas under the
normal curve. The incubation times that make up the middle 97% are 10 days. (Round to the
nearest whole number ass needed. Use ascending order.).
ind the area of the shaded region The graph to the right de.pdf
ind the area of the shaded region. The graph to the right depicts 1Q scores of adults, and those
scores are normally listributed with a mean of 100 and a slandard deviation of 15 . The atea of the
shaded region is (fround to four decimal placos as mooded).
increase in the demand for loanable funds could be best expl.pdf
increase in the demand for loanable funds could be best explained by which of the following? (A)
There is a decrease in investment spending. (B) There is an increase in the government's budget
surplus. (C) Firms are optimistic about the future performance of the country's economy. (D)
Domestic investors seek higher returns by investing in foreign financial assets. (E) The economy is
facing political instability..
Increase the proportion of aduits who get recommended eviden.pdf
Increase the proportion of aduits who get recommended evidence-based preventive healthcare
(AHS-08) Increase the number of community organizations that provide prevention services
(ECBP-D07) Reduce the proportion of people who can't get prescription medicines when they
need them (AHS-06) Increase the proportion of people with a usual primary care provider (AHS-
07) Increase the proportion of people with prescription drug insurance (AHS-03)For each
healthcare access and quality-related objective, identify if the objective is a core, research, or
developmental objective..
In your team research how you can improve communications in.pdf
In your team, research how you can improve communications in global teams. Provide the results
of your research in a not-to-exceed two-page double-spaced paper. Include in your discussion, as
a minimum: (a) How managing communications in agile is different? (b) What are the best
practices for improving communication in global teams?.
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..
include ltiostreamgt include ltfstreamgt in.pdf
#include <iostream>
#include <fstream>
#include <sstream>
#include <vector>
#include <string>
using namespace std;
/* A binary tree node has data,pointer to left child and apointer to right child */
struct Node
{
int data;
Node* left, * right;
};
/*Function for creating a new Node structure */
Node* newNode(int data)
{
Node* node = new Node;
//Node* node = (Node*)malloc(sizeof(Node)); //will also work.
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Function to insert nodes in level order. This function takes an array-based treerepresentation as
inputand
// convert it into a linked list-based tree representationNode* insertLevelOrder(vector<int>
tree_vec,
//tree_vec: array implementation,
create node for position i.
int i, int n)
{
Node* root = nullptr;
// Base case for recursion
if (i < n)
{
root = newNode(tree_vec[i]);
// insert left child
root->left = insertLevelOrder(tree_vec,
2 * i + 1, n);
// insert right child
root->right = insertLevelOrder(tree_vec,
2 * i + 2, n);
}
return root;
}
// Example code for InOrder traversal
void inOrder(Node * root)
{
if (root != NULL)
{
inOrder(root->left);
cout << root->data << " ";
inOrder(root->right);
}
}
int main()
{
vector<int> tree_vec;
string file_name;
int temp_data;
cout << "Please input the file name of a complete binary tree:";
cin >> file_name;
ifstream tree_file(file_name, ios::in);
if (tree_file.fail())
{
cout << "Fail to open the tree file.";
return 0;
}
while (!tree_file.fail() && !tree_file.eof())
{
tree_file >> temp_data;
tree_vec.push_back(temp_data);
}
/*Transform the array-based tree into linked-list in level-order.*/
Node* root = insertLevelOrder(tree_vec, 0, tree_vec.size());
/*Task 1, to be implemented*/
cout << "The Preorder Traversal result is: " << endl;
/*Example implementation*/
cout << "The Inorder Traversal result is: " << endl;
inOrder(root);
/*Task 2, to be implemented*/
cout << "The Postorder Traversal result is: " << endl;
/*Task 3, to be implemented*/
cout << "The number of leaf nodes is: " << endl;
return 0;
}
please help me to finish the code with the task using c++
#include <iostream>
#include <fstream>
#include <sstream>
#include <vector>
#include <string>
using namespace std;
/* A binary tree node has data,pointer to left child and apointer to right child */
struct Node
{
int data;
Node* left, * right;
};
/*Function for creating a new Node structure */
Node* newNode(int data)
{
Node* node = new Node;
//Node* node = (Node*)malloc(sizeof(Node)); //will also work.
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Function to insert nodes in level order. This function takes an array-based treerepresentation as
inputand
// convert it into a linked list-based tree representationNode* insertLevelOrder(vector<int>
tree_vec,
//tree_vec: array implementation,
create node for position i.
int i, int n)
{
Node* root = nullptr;
// Base case for recursion
if (i < n)
{
root = newNode(tree_vec[i]);
// insert left child
root->left = inse.
include ltstdiohgtinclude ltstdlibhgtinclude l.pdf
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void FillBingoCard(int bingocard[5][5]);
void PrintCard(int bingocard[5][5], int called[3], char yn[1]);
int PickNumber(int called[3], int callednumber[75], int counter);
int drawnnumber(int bingocard[5][5], int called[3]);
void completedrowcolumn(int bingocard[5][5], int win);
int main(void)
{
//array of bingo numbers
int row = 5;
int col = 5;
int bingocard[row][col];
int called[3];
int counter = 0;
int callednumber[75];
char yn[1];
int win = 0;
//fill bingocard
void FillBingoCard(int bingocard[5][5])
{
int i, j, k;
int temp;
srand(time(NULL));
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
temp = rand() % 15 + 1 + j * 15;
for (k = 0; k < j; k++)
{
if (temp == bingocard[i][k])
{
j--;
break;
}
}
bingocard[i][j] = temp;
}
}
bingocard[2][2] = -1;
}
void PrintCard(int bingocard[5][5], int called[3], char yn[1])
{
int i, j;
printf(" B I N G On");
for (i = 0; i < 5; i++)
{
printf("----------------------n");
for (j = 0; j < 5; j++)
{
if (bingocard[i][j] == -1)
{
printf("| X ");
}
else if (called[0] == 'B' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else if (called[0] == 'I' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else if (called[0] == 'N' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else if (called[0] == 'G' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else if (called[0] == 'O' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else
{
printf("| %2d", bingocard[i][j]);
}
}
printf("|n");
}
printf("----------------------n");
printf("Have you got number %c%d? (Y/N) ", called[0], called[1]);
scanf(" %c", yn);
}
int PickNumber(int called[3], int callednumber[75], int counter)
{
int temp;
srand(time(NULL));
do
{
temp = rand() % 75 + 1;
} while (callednumber[temp - 1] != 0);
callednumber[temp - 1] = 1;
called[1] = temp;
if (temp <= 15)
{
called[0] = 'B';
}
else if (temp <= 30)
{
called[0] = 'I';
}
else if (temp <= 45)
{
called[0] = 'N';
}
else if (temp <= 60)
{
called[0] = 'G';
}
else
{
called[0] = 'O';
}
return *called;
}
int drawnnumber(int bingocard[5][5], int called[3])
{
int row, col;
int found = 0;
for (row = 0; row < 5; row++)
{
for (col = 0; col < 5; col++)
{
if (bingocard[row][col] == called[1])
{
bingocard[row][col] = -1;
found = 1;
}
}
}
if (found == 1)
{
return 1;
}
else
{
return 0;
}
}
{
int row, col;
// check for completed rows
for (row = 0; row < 5; row++)
{
int count = 0;
for (col = 0; col < 5; col++)
{
if (bingocard[row][col] == -1)
{
count++;
}
}
if (count == 5)
{
printf("nBINGO! You completed row %d.", row + 1);
win = 1;
}
}
// check for completed columns
for (col = 0; col < 5; col++)
{
int count = 0;
for (row = 0; row < 5; row++)
{
if (bingocard[row][col] == -1)
{
count++;
}
}
if (count == 5)
{
printf("nBINGO! You completed column %d.", col + 1);
win = 1;
}
}
}
void completedrowcolumn(int bingocard[5][5], int win)
{
int row, col, diag, count;
// ch.
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptxMohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
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template<typename T>
struct ListNode
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In your team research how you can improve communications in.pdf
In your team, research how you can improve communications in global teams. Provide the results
of your research in a not-to-exceed two-page double-spaced paper. Include in your discussion, as
a minimum: (a) How managing communications in agile is different? (b) What are the best
practices for improving communication in global teams?.
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..
include ltiostreamgt include ltfstreamgt in.pdf
#include <iostream>
#include <fstream>
#include <sstream>
#include <vector>
#include <string>
using namespace std;
/* A binary tree node has data,pointer to left child and apointer to right child */
struct Node
{
int data;
Node* left, * right;
};
/*Function for creating a new Node structure */
Node* newNode(int data)
{
Node* node = new Node;
//Node* node = (Node*)malloc(sizeof(Node)); //will also work.
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Function to insert nodes in level order. This function takes an array-based treerepresentation as
inputand
// convert it into a linked list-based tree representationNode* insertLevelOrder(vector<int>
tree_vec,
//tree_vec: array implementation,
create node for position i.
int i, int n)
{
Node* root = nullptr;
// Base case for recursion
if (i < n)
{
root = newNode(tree_vec[i]);
// insert left child
root->left = insertLevelOrder(tree_vec,
2 * i + 1, n);
// insert right child
root->right = insertLevelOrder(tree_vec,
2 * i + 2, n);
}
return root;
}
// Example code for InOrder traversal
void inOrder(Node * root)
{
if (root != NULL)
{
inOrder(root->left);
cout << root->data << " ";
inOrder(root->right);
}
}
int main()
{
vector<int> tree_vec;
string file_name;
int temp_data;
cout << "Please input the file name of a complete binary tree:";
cin >> file_name;
ifstream tree_file(file_name, ios::in);
if (tree_file.fail())
{
cout << "Fail to open the tree file.";
return 0;
}
while (!tree_file.fail() && !tree_file.eof())
{
tree_file >> temp_data;
tree_vec.push_back(temp_data);
}
/*Transform the array-based tree into linked-list in level-order.*/
Node* root = insertLevelOrder(tree_vec, 0, tree_vec.size());
/*Task 1, to be implemented*/
cout << "The Preorder Traversal result is: " << endl;
/*Example implementation*/
cout << "The Inorder Traversal result is: " << endl;
inOrder(root);
/*Task 2, to be implemented*/
cout << "The Postorder Traversal result is: " << endl;
/*Task 3, to be implemented*/
cout << "The number of leaf nodes is: " << endl;
return 0;
}
please help me to finish the code with the task using c++
#include <iostream>
#include <fstream>
#include <sstream>
#include <vector>
#include <string>
using namespace std;
/* A binary tree node has data,pointer to left child and apointer to right child */
struct Node
{
int data;
Node* left, * right;
};
/*Function for creating a new Node structure */
Node* newNode(int data)
{
Node* node = new Node;
//Node* node = (Node*)malloc(sizeof(Node)); //will also work.
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Function to insert nodes in level order. This function takes an array-based treerepresentation as
inputand
// convert it into a linked list-based tree representationNode* insertLevelOrder(vector<int>
tree_vec,
//tree_vec: array implementation,
create node for position i.
int i, int n)
{
Node* root = nullptr;
// Base case for recursion
if (i < n)
{
root = newNode(tree_vec[i]);
// insert left child
root->left = inse.
include ltstdiohgtinclude ltstdlibhgtinclude l.pdf
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void FillBingoCard(int bingocard[5][5]);
void PrintCard(int bingocard[5][5], int called[3], char yn[1]);
int PickNumber(int called[3], int callednumber[75], int counter);
int drawnnumber(int bingocard[5][5], int called[3]);
void completedrowcolumn(int bingocard[5][5], int win);
int main(void)
{
//array of bingo numbers
int row = 5;
int col = 5;
int bingocard[row][col];
int called[3];
int counter = 0;
int callednumber[75];
char yn[1];
int win = 0;
//fill bingocard
void FillBingoCard(int bingocard[5][5])
{
int i, j, k;
int temp;
srand(time(NULL));
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
temp = rand() % 15 + 1 + j * 15;
for (k = 0; k < j; k++)
{
if (temp == bingocard[i][k])
{
j--;
break;
}
}
bingocard[i][j] = temp;
}
}
bingocard[2][2] = -1;
}
void PrintCard(int bingocard[5][5], int called[3], char yn[1])
{
int i, j;
printf(" B I N G On");
for (i = 0; i < 5; i++)
{
printf("----------------------n");
for (j = 0; j < 5; j++)
{
if (bingocard[i][j] == -1)
{
printf("| X ");
}
else if (called[0] == 'B' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else if (called[0] == 'I' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else if (called[0] == 'N' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else if (called[0] == 'G' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else if (called[0] == 'O' && called[1] == bingocard[i][j])
{
printf("| %2d", bingocard[i][j]);
}
else
{
printf("| %2d", bingocard[i][j]);
}
}
printf("|n");
}
printf("----------------------n");
printf("Have you got number %c%d? (Y/N) ", called[0], called[1]);
scanf(" %c", yn);
}
int PickNumber(int called[3], int callednumber[75], int counter)
{
int temp;
srand(time(NULL));
do
{
temp = rand() % 75 + 1;
} while (callednumber[temp - 1] != 0);
callednumber[temp - 1] = 1;
called[1] = temp;
if (temp <= 15)
{
called[0] = 'B';
}
else if (temp <= 30)
{
called[0] = 'I';
}
else if (temp <= 45)
{
called[0] = 'N';
}
else if (temp <= 60)
{
called[0] = 'G';
}
else
{
called[0] = 'O';
}
return *called;
}
int drawnnumber(int bingocard[5][5], int called[3])
{
int row, col;
int found = 0;
for (row = 0; row < 5; row++)
{
for (col = 0; col < 5; col++)
{
if (bingocard[row][col] == called[1])
{
bingocard[row][col] = -1;
found = 1;
}
}
}
if (found == 1)
{
return 1;
}
else
{
return 0;
}
}
{
int row, col;
// check for completed rows
for (row = 0; row < 5; row++)
{
int count = 0;
for (col = 0; col < 5; col++)
{
if (bingocard[row][col] == -1)
{
count++;
}
}
if (count == 5)
{
printf("nBINGO! You completed row %d.", row + 1);
win = 1;
}
}
// check for completed columns
for (col = 0; col < 5; col++)
{
int count = 0;
for (row = 0; row < 5; row++)
{
if (bingocard[row][col] == -1)
{
count++;
}
}
if (count == 5)
{
printf("nBINGO! You completed column %d.", col + 1);
win = 1;
}
}
}
void completedrowcolumn(int bingocard[5][5], int win)
{
int row, col, diag, count;
// ch.
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