SOLUTION- CHAPTER-1: ( Over view of C ) from PROGRAMMING IN ANSI C: E BALA GURUSHAMY

SOLUTION- CHAPTER-1: ( Over view of C )

PROGRAMMING IN ANSI C: E BALA GURUSHAMY

Problem exercise no 1.1&1.2:

Coding of the programme:

#include

#include

void main()

{

clrscr();

printf("---------------------------------------\n");

printf("I First line :A.Z.M.Shakilur Rahman I\nI Second line :12/a ,Ali sonar lane I\nI Third line:Bogra,5800 I\n");

printf("---------------------------------------");

getch();

}

Output:

Problem exercise no. 1.3:

Coding of the programme:

#include

#include

void main()

{clrscr();

printf("*\n* *\n* * *\n* * * * ");

getch();

}

Output:

*

* *

* * *

* * * *

Problem exercise no :1.4

Coding of the problem:

#include

#include

void main()

{clrscr();

printf("a>>------------------>b");

getch();

}

Output:

a>>------------------>b

Problem exercise no:1.5

Coding of the problem:

#include

#include

#define pi 3.14159

void main()

{

float r,A;

clrscr();

printf("\n\tENTER THE RADIUS OF A CIRCLE=");

scanf("%f",&r);

A=pi*r*r;

printf("\n\n\tArea=%f sqr unit",A);

getch();

}

Output:

ENTER THE RADIUS OF A CIRCLE=2

Area=12.566360 sqr unit

Problem exercise no:1.6

CODING:

#include

#include

void main()

{

int b,c;

clrscr();

for(b=1;b<=10;b++)

{

c=5*b;

printf("\n\t%d*%d=%d\n",5,b,c);

getch();

}

}

Output :

Problem exercise no:1.7

Coding of the programme:

#include

#include

void add();

void sub();

void main()

{

clrscr();

add();

sub();

getch();

}

void add()

{

printf("\n\t%d+%d=%d",20,10,30);

}

void sub()

{

printf("\n\t%d-%d=%d",20,10,10);

}

Output :

20+10=30

20-10=10

Problem exercise no:1.8

Coding:

#include

#include

void main()

{

int a,b,c,x;

clrscr();

printf("Enter values of a,b&c\n");

scanf("%d%d%d",&a,&b,&c);

x=a/(b-c);

printf("result=%d",x);

getch();

}

Output:

a)

Enter values of a,b&c

250

85

25

result=4

b)NO OUTPUT

Problem exercise no:1.9 (b)

Coding :

#include

#include

void main()

{

float a,F,C;

clrscr();

printf("ENTER TEMPERATURE IN FARENHITE\n");

scanf("%f",&F);

a=5*(F-32);

C=a/9;

printf("\nIn celsius scale=%f",C);

getch();

}

Output :

ENTER TEMPERATURE IN FARENHITE

10

In Celsius scale=-12.222222

Problem exercise no:1.9 (a)

Coding :

#include

#include

void main()

{

float a,F,C;

clrscr();

printf("ENTER TEMPERATURE IN CELSIUS\n");

scanf("%f",&C);

a=(9*C)/5;

F=a+32;

printf("\nIn farenhite scale=%f",F);

getch();

}

Output:

ENTER TEMPERATURE IN CELSIUS

10

In frenhite scale=50.00000

Problem exercise no: 1.10

Coding of the problem:

#include

#include

#include

void main()

{

clrscr();

float a,b,c,S,A;

printf("\n\tENTER THE THREE SIDES OF A TRIANGLE=");

scanf("%f%f%f",&a,&b,&c);

S=(a+b+c)/2;

A=sqrt(S*(S-a)*(S-b)*(S-c));

printf("\n\tArea of the triangle=%f",A);

getch();

}

Sample output:

ENTER THE THREE SIDES OF A TRIANGLE=10

12

14

Area of the triangle=58.787754

Problem exercise no:1.11

Coding:

#include

#include

#include

void main()

{

float D,x1,x2,y1,y2;

clrscr();

printf("ENTER CO-ORDINATES x1,x2,y1,y2=\n");

scanf("%f%f%f%f",&x1,&x2,&y1,&y2);

D=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));

printf("Result=%f",D);

getch();

}

Output :

ENTER CO-ORDINATES x1,x2,y1,y2=

2

4

8

5

Result=3.605551

Problem exercise no:1.12

Coding:

#include

#include

#include

#define pi 3.14159

void main()

{

float r,x1,x2,y1,y2,A;

clrscr();

x1=0;

x2=0;

y1=4;

y2=5;

r=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));

A=pi*r*r;

printf("Result=%f",A);

getch();

}

Output :

Result=3.14159

Problem exercise no:1.13

Coding:

#include

#include

#include

#define pi 3.14159

void main()

{

float D,r,x1,x2,y1,y2,A;

clrscr();

x1=2;

x2=2;

y1=5;

y2=6;

D=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));

r=D/2;

A=pi*r*r;

printf("Result=%f",A);

getch();

}

Output :

Result=0.785398

Problem exercise no:1.14

Coding:

#include

#include

void main()

{ int a,b,c;

clrscr();

a=5;

b=8;

c=18;

printf("%dx+%dy=%d",a,b,c);

getch();

}

Output :

5x+8y=18

Problem exercise no:1.15

Coding:

#include

#include

void main()

{ float x,y,sum,difference,product,division;

clrscr();

printf("ENTER TWO NUMBERS=\n");

scanf("%f%f",&x,&y);

sum=x+y;

difference=x-y;

product=x*y;

division=x/y;

printf("\n\tSum=%f\tDifference=%f\n\n\tProduct=%f\tDivision=%f",sum,difference,product,division);

getch();

}

Output :

ENTER TWO NUMBERS=

10

5

Sum=15.000000 Difference=5.000000

Product=50.000000 Division=2.000000