SPL 6.1 | Advanced problems on Operators and Math.h function in C

Structured Programming Language
# include <math.h>
Mohammad Imam Hossain,
Lecturer, CSE, UIU
Advanced problems on Operators

Flow Chart
• This whole slide is based on the following flow chart:

# include <math.h>
The math.h header declares a set of functions
to compute common mathematical operations
and transformations.
Ref: MathDotH.pdf
“ Computers are good
at following
instructions, but not at
reading your mind. ” -
Donald Knuth

Problem
• Input: x (double)
• Output: y (double)
𝑦 = 5𝑥4 + 3𝑥2 + 7 tan 𝑥 − 𝑒 𝑥 + 𝑥

Solution
𝑦 = 5𝑥4 + 3𝑥2 + 7 tan 𝑥 − 𝑒 𝑥 + 𝑥#include <stdio.h>
int main()
{
return 0;
}

Solution
int main()
{
double x,y; ///variable declaration
return 0;
}

Solution
int main()
{
printf("Please enter the value of x: ");
scanf(" %lf",&x); ///taking input the value of x
return 0;
}

Solution
#include <math.h>
int main()
{
y=5*pow(x,4)+3*x*x+sqrt(7)*tan(x)-exp(x)+ceil(fabs(x)); ///Processing
return 0;
}

Solution
#include <math.h>
int main()
{
y=5*pow(x,4)+3*x*x+sqrt(7)*tan(x)-exp(x)+ceil(fabs(x)); ///Processing
printf("The resulting value is %lfn",y); ///printing output
return 0;
}

Practice
• Input: a (double),
b(double) ,
c(double)
• Output: x (double)
𝑥 =
5𝑎2
+ 6𝑏 + 𝑐
2𝑎 + 𝑐

Problem
• Input: p (integer, no of watts of a light),
t (double, no of hours it is turned on)
• Output: no of B.O.T units in kwh format.

Problem
• Input: p (integer, no of watts of a light),
t (double, no of hours it is turned on)
• Output: no of B.O.T units in kwh format.
units =
𝑝𝑡
1000
𝐾𝑊𝐻

Code Sample
#include <stdio.h>
int main()
{
return 0;
}

Code Sample
#include <stdio.h>
int main()
{
double p,t,result;
printf("Enter the power(in watts) & used time(in hours): ");
scanf("%lf %lf",&p,&t); ///taking input
return 0;
}

Code Sample
#include <stdio.h>
int main()
{
double p,t,result;
result=p*t/1000; ///calculating no of units
return 0;
}

Code Sample
#include <stdio.h>
int main()
{
double p,t,result;
result=p*t/1000; ///calculating no of units
printf("No of units: %.2lfn",result); ///output
return 0;
}

Problem
• Write a program that will initialize two integer types
of variables (a ,b) and interchange(swap) the values
between them and finally show the output.
• Input: a=10, b=20
• Output: a=20, b=10

Code Sample
#include <stdio.h>
int main()
{
int a,b;
scanf("%d %d",&a,&b); ///taking inputs
return 0;
}

Code Sample
#include <stdio.h>
int main()
{
int a,b;
int temp; /// to hold temporary value
temp=a; /// saving a into temp
return 0;
}

Code Sample
#include <stdio.h>
int main()
{
int a,b;
a=b; ///now a has the new value of b
return 0;
}

Code Sample
#include <stdio.h>
int main()
{
int a,b;
b=temp; ///now b has the value of temp(old value of a)
return 0;
}

Code Sample
#include <stdio.h>
int main()
{
int a,b;
b=temp; ///now b has the value of temp(old value of a)
printf("a=%d and b=%dn",a,b); ///output
return 0;
}

Problem
• Input: temperature(double) in °C format.
• Output: temperature(double) in °F format.

Problem
• Input: temperature(double) in °C format.
• Output: temperature(double) in °F format.
𝐶 − 0
100 − 0
=
𝐹 − 32
212 − 32

Code Sample
#include <stdio.h>
int main()
{
double c,f;
printf("Enter temperature in Celsius: ");
scanf("%lf",&c); ///taking input Celsius scale temperature
return 0;
}

Code Sample
#include <stdio.h>
int main()
{
double c,f;
f= 9*c/5+32; /// processing
return 0;
}

Code Sample
#include <stdio.h>
int main()
{
double c,f;
f= 9*c/5+32; /// processing
printf("%.2lf Celsius = %.2lf Fahrenheit",c,f); ///output
return 0;
}

Problem
• Input: r (double, radius),
h (double, height)
• Output: the volume of a sphere

Problem
• Input: r (double, radius),
h (double, height)
• Output: the volume of a sphere
𝜋𝑟2ℎ

Code Sample
#include <stdio.h>
int main()
{
double radius,height,volume;
printf("please enter the radius and height of a cylinder: ");
scanf("%lf %lf",&radius,&height); ///taking inputs
return 0;
}

Code Sample
#include <stdio.h>
#define PI 3.1416
int main()
{
volume = PI*radius*radius*height; ///calculating volume
return 0;
}

Code Sample
#include <stdio.h>
#define PI 3.1416
int main()
{
volume = PI*radius*radius*height; ///calculating volume
printf("The volume of cylinder is %.3lfn",volume); ///output
return 0;
}

Problem
• Input: 6 floating point numbers
(x1, y1),
(x2, y2),
(x3, y3)
• Output: area (double, area of a triangle)
(𝑥1, 𝑦1)
(𝑥2, 𝑦2) (𝑥3, 𝑦3)

Problem
• Input: 6 floating point numbers
(x1, y1),
(x2, y2),
(x3, y3)
• Output: area (double, area of a triangle)
(𝑥1, 𝑦1)
(𝑥2, 𝑦2) (𝑥3, 𝑦3)
𝑎𝑟𝑒𝑎 =
1
2
𝑥1 − 𝑥2 ∗ 𝑦2 − 𝑦3 − 𝑦1 − 𝑦2 ∗ 𝑥2 − 𝑥3
=
1
2
[𝑥1 ∗ 𝑦2 − 𝑦3 + 𝑥2 ∗ 𝑦3 − 𝑦1 + 𝑥3 ∗ 𝑦1 − 𝑦2 ]

Code Sample#include <stdio.h>
int main()
{
return 0;
}

int main()
{
double x1,y1,x2,y2,x3,y3;
printf("Enter the 3 vertices of a triangle: ");
scanf("%lf %lf",&x1,&y1);
scanf("%lf %lf",&x3,&y3); ///taking input
return 0;
}

#include <math.h> ///for function fabs
int main()
{
double area;
area=0.5*(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2));///calculating
return 0;
}

#include <math.h> ///for function fabs
int main()
{
double area;
area=0.5*(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2));///calculating
printf("The area is %.3lf sq units.n“,fabs(area));///outputs
return 0;
}

Practice
• Input: a , b (base and height of a right angle triangle)
• Output: the value of the hypotenuse

Practice
• Input: a , b (base and height of a right angle triangle)
• Output: the value of the hypotenuse
ℎ𝑦𝑝 = 𝑏𝑎𝑠𝑒2 + ℎ𝑒𝑖𝑔ℎ𝑡2
base
hypotenuse
height

Problem
• Input: t (integer, time in seconds)
• Output: ddd days hh hours mm minutes and ss seconds
• Sample I/O:
Input: 123456
Output: 001 days 10 hours 17 minutes and 36 seconds
“ First, solve the
problem. Then,
write the code. ” -
John Johnson

int main()
{
int input;
scanf("%d",&input); ///taking input
return 0;
}

int main()
{
int input;
int day,hr,min,sec,rem_sec;
return 0;
}

int main()
{
int input;
day=input/(24*60*60); ///calculating no of days
rem_sec=input%(24*60*60); ///the remaining seconds
return 0;
}

int main()
{
int input;
hr=rem_sec/(60*60); ///calculating no of hours
rem_sec=rem_sec%(60*60); ///remaining seconds
return 0;
}

int main()
{
int input;
min=rem_sec/60; ///calculating no of minutes
sec=rem_sec%60; ///final remaining seconds
return 0;
}

int main()
{
int input;
min=rem_sec/60; ///calculating no of minutes
sec=rem_sec%60; ///final remaining seconds
printf("%03d days %02d hours %02d minutes and %02d seconds",day,hr,min,sec); ///output
return 0;
}

Problem
• Input: an integer of exactly 4 digits (for example, 1234)
• Output: 1 thousands, 2 hundreds and 34

int main()
{
int num;
scanf("%4d",&num); ///taking input
return 0;
}

int main()
{
int num;
int thous,hund,rem,temp; ///declaring variables
return 0;
}

int main()
{
int num;
thous=num/1000; /// calculating thousands
temp=num%1000; /// remaining amounts
return 0;
}

int main()
{
int num;
hund=temp/100; ///calculating hundreds
rem=temp%100; ///final remaining amounts
return 0;
}

int main()
{
int num;
hund=temp/100; ///calculating hundreds
rem=temp%100; ///final remaining amounts
printf("%d thousands, %d hundreds and %dn",thous,hund,rem); ///output
return 0;
}

Practice
• Input: an integer of exactly 4 digits (let, 1234)
• Output: the sum of the digits (10)

Problem
• Input: a, b, c of the equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
• Output: the roots of this equation.

Problem
• Input: a, b, c of the equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
• Output: the roots of this equation.
𝑥1 =
−𝑏 + 𝑏2 − 4𝑎𝑐
2𝑎
𝑥2 =
−𝑏 − 𝑏2 − 4𝑎𝑐
2𝑎

Code Sample
#include <stdio.h>
#include <math.h>
int main()
{
return 0;
}

Code Sample
#include <stdio.h>
#include <math.h>
int main()
{
int a,b,c; ///ax^2+bx+c=0, coefficient input
scanf("%d %d %d",&a,&b,&c);
return 0;
}

Code Sample
#include <stdio.h>
#include <math.h>
int main()
{
double x1,x2; ///roots calculation
x1=(-b+sqrt(b*b-4*a*c))/(2*a);
x2=(-b-sqrt(b*b-4*a*c))/(2*a);
return 0;
}

Code Sample
#include <stdio.h>
#include <math.h>
int main()
{
double x1,x2; ///roots calculation
x1=(-b+sqrt(b*b-4*a*c))/(2*a);
x2=(-b-sqrt(b*b-4*a*c))/(2*a);
printf("Solutions are %lf and %lfn",x1,x2); ///output
return 0;
}

Practice
• Input: a1, b1, c1 and a2, b2, c2
where,
𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0
𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0
• Output: the intersection point of that two straight
lines.

Practice
• Input: a1, b1, c1 and a2, b2, c2
where,
𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0
𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0
• Output: the intersection point of that two straight
lines.
𝑥 =
𝑏1 𝑐2 − 𝑏2 𝑐1
𝑎1 𝑏2 − 𝑎2 𝑏1
𝑦 =
𝑐1 𝑎2 − 𝑐2 𝑎1
𝑎1 𝑏2 − 𝑎2 𝑏1

Practice
• Input: two integers indicating the hours and minutes
of an analog clock
• Output: the angle between the two indicators.

Problem
• Input: 3 angles(double) of a triangle.
• Output: a character
‘Y’ (if valid triangle)
‘N’ (if not a valid triangle)
A
B C
A+B+C = 𝜋

Code Sample
#include <stdio.h>
int main()
{
double A,B,C;
scanf("%lf %lf %lf",&A,&B,&C);
char result;
result=(A+B+C-180.0 < 0.00000001)?'Y':'N';
printf("%cn",result);
return 0;
}

Practice
• Input: 3 double numbers from the user
• Output: average (double) of that 3 numbers.
Sample I/O:
• Input: 10.0,20.0,30.0
• Output: 20.000000

Practice
• Input: Grade points(double) of 6 subjects and their
corresponding credit hours(double).
• Output: weighted GPA
𝐺𝑃𝐴 =
σ𝑖=1
𝑛
𝐶𝑖 ∗ 𝐺𝑖
σ𝑖=1
𝑛
𝐶𝑖

Course Credits, 𝑪𝒊 Grade Points, 𝑮𝒊 𝑪𝒊 * 𝑮𝒊
CSE 100 2.00 4.00 8.00
EEE 163 3.00 4.00 12.00
EEE 164 1.50 3.75 5.625
MATH 141 3.00 3.00 9.00
ME 160 1.50 3.50 5.250
ME 165 3.00 4.00 12.00
PHY 109 4.00 3.75 15.00
PHY 102 1.50 3.50 5.250
Total 19.50 72.125
GPA calculation
𝐺𝑃𝐴 =
72.125
19.50
= 3.7

Practice
• Input: Total credits(double) of 4 semesters and their
corresponding GPA(double) earned.
• Output: Cumulative GPA
C𝐺𝑃𝐴 =
σ𝑖=1
𝑛
𝑇𝐶𝑖 ∗ 𝐺𝑃𝐴𝑖
σ𝑖=1
𝑛
𝑇𝐶𝑖

Semester Total Credits, 𝐓𝑪𝒊 GPA earned, 𝑮𝑷𝑨𝒊 𝑻𝑪𝒊 * 𝑮𝑷𝑨𝒊
1 19.50 3.70 72.150
2 20.50 3.93 80.565
3 21.25 3.96 84.150
4 20.25 4.00 81.00
Total 81.50 317.865
CGPA calculation
𝐶𝐺𝑃𝐴 =
317.865
81.50
= 3.90

SPL 6.1 | Advanced problems on Operators and Math.h function in C

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