SEBA class 10 computer science | chapter 5 : Nested loops in C | exercise and additional questions answers
SEBA class 10 computer science | chapter 5: Nested loops in C | exercise and additional questions answers are given below.
Exercise :
- Write C program to display the following patterns using nested loop construct.
a. 1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
Ans: #include<stdio.h>
int main()
{
int i,j;
for(i=1;i<=5;i++)
{
for(j=1;j<=3;j++)
{
printf(“%d\t”,j);
}
printf(“\n”);
}
return 0;
}
b. 1 2 1
1 2 1
1 2 1
1 2 1
1 2 1
Ans:
#include<stdio.h>
int main()
{
int i,j;
for(i=1;i<=5;i++)
{
for(j=1;j<=2;j++)
{
printf(“%d\t”,j);
}
printf(“1\t”);
printf(“\n”);
}
return 0;
}
c. 4 3 2 1
4 3 2 1
4 3 2 1
4 3 2 1
4 3 2 1
Ans: #include<stdio.h>
int main()
{
int i,j;
for(i=1;i<=5;i++) { for(j=4;j>=1;j–)
{
printf(“%d\t”,j);
}
printf(“\n”);
}
return 0;
}
d. 2
2 3 4
2 3 4 5 6
Ans: #include<stdio.h>
int main()
{
int i,j,k,m;
for(i=1;i<=3;i++)
{
for(j=1;j<=(3-i);j++)
{
printf(” \t”);
}
for(k=2;k<=2*i;k++)
{
printf(“%d\t”,k);
}
for(m=1;m<=(3-i);m++)
{
printf(” \t”);
}
printf(“\n”);
}
return 0;
}
e. 1
1 2 1
1 2 3 2 1
Ans: #include<stdio.h>
int main()
{
int i,j,m,k;
for(i=1;i<=3;i++)
{
for(j=1;j<=(3-i);j++)
{
printf(" \t");
}
for(m=1;m<=i;m++)
{printf(“%d\t”,m);
}
for(k=(i-1);k>=1;k–)
{
printf(“%d\t”,k);
}
printf(“\n”);
}
return 0;
}
f.
*
* * *
* * * * *
* * * * * * *
* * * * *
* * *
*
Ans: #include<stdio.h>
int main()
{
int i,j,k,m,n,x;
for(i=1;i<=4;i++)
{
for(j=1;j<=(4-i);j++)
{
printf(” \t”);
}
for(k=1;k<=(2*i-1);k++)
{
printf(“X\t”);
}
printf(“\n”);
}
for(x=1;x<=3;x++)
{
for(m=1;m<=x;m++)
{
printf(” \t”);
}
for(n=7;n>=(x*2+1);n–)
{
printf(“X\t”);
}
printf(“\n”);
}
return 0;
}
g.
* *
* * * *
* * * * * *
* * * * * * * *
* * * * * * * * * *
Ans: #include<stdio.h>
int main()
{
int i,j,k,m,n;
for(i=1;i<=5;i++)
{
for(j=1;j<=i;j++)
{
printf(“X”);
}
for(k=4;k>=1;k–)
{
printf(” “);
}
for(j=1;j<=i;j++)
{
printf(“X”);
}
for(k=4;k>=1;k–)
{
printf(” “);
}
printf(“\n”);
}
return 0;
}
2. Modify the solution of question no. 1 to accept the number of lines as the input. The program should make the display pattern accordingly ( hint : Write separate programs)
Ans: 1. a. #include<stdio.h>
int main()
{
int n,i,j;
printf(“Enter the number of lines to be printed”);
scanf(“%d”,&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=3;j++)
{
printf(“%d\t”,j);
}
printf(“\n”);
}
return 0;
}
output:
- b. #include<stdio.h>
int main()
{
int n,i,j;
printf(“Enter the number of lines to be printed”);
scanf(“%d”,&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=2;j++)
{
printf(“%d\t”,j);
}
printf(“1\t”);
printf(“\n”);
}
return 0;
}
Output :
- c. #include<stdio.h>
int main()
{
int n,i,j;
printf(“Enter the number of lines to be printed”);
scanf(“%d”,&n);
for(i=1;i<=n;i++) { for(j=4;j>=1;j–)
{
printf(“%d\t”,j);
}
printf(“\n”);
}
return 0;
}
Output :
- d. #include<stdio.h>
int main()
{
int n,i,j,k,m;
printf(“Enter the number of lines to be printed”);
scanf(“%d”,&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=(n-i);j++)
{
printf(” \t”);
}
for(k=2;k<=2*i;k++)
{
printf(“%d\t”,k);
}
for(m=1;m<=(3-i);m++)
{
printf(” \t”);
}
printf(“\n”);
}
return 0;
}
Output :
- e. #include<stdio.h>
int main()
{
int n,i,j,m,k;
printf(“Enter the number of lines to be printed”);
scanf(“%d”,&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=(3-i);j++)
{
printf(” \t”);
}
for(m=1;m<=i;m++)
{
printf("%d\t",m);}
for(k=(i-1);k>=1;k–)
{
printf(“%d\t”,k);
}
printf(“\n”);
}
return 0;
}
Output:
- f. #include<stdio.h>
int main()
{
int a,b,i,j,k,m,n,x;
printf(“Enter the number of lines to be printed in the top pattern”);
scanf(“%d”,&a);
printf(“Enter the number of lines to be printed in the bottom pattern”);
scanf(“%d”,&b);
for(i=1;i<=a;i++)
{
for(j=1;j<=(4-i);j++)
{
printf(” \t”);
}
for(k=1;k<=(2*i-1);k++)
{
printf(“X\t”);
}
printf(“\n”);
}
for(x=1;x<=b;x++)
{
for(m=1;m<=x;m++)
{
printf(” \t”);
}
for(n=7;n>=(x*2+1);n–)
{
printf(“X\t”);
}
printf(“\n”);
}
return 0;
}
Output:
g. #include<stdio.h.int main()
{
int a,i,j,k,m,n;
printf(“Enter the number of lines to be printed”);
scanf(“%d”,&a);
for(i=1;i<=a;i++)
{
for(j=1;j<=i;j++)
{
printf(“X”);
}
for(k=4;k>=1;k–)
{
printf(” “);
}
for(j=1;j<=i;j++)
{
printf(“X”);
}
for(k=4;k>=1;k–)
{
printf(” “);
}
printf(“\n”);
}
return 0;
}
Output :
Enter the number of lines to be printed 5
X X
XX XX
XXX XXX
XXXX XXXX
XXXXX XXXXX
Process returned 0 (0x0) execution time : 2.537 s
Press any key to continue.
Related Posts
Hello students,
Welcome to my site. I am the founder of the blog elearnersmentor.com. I have been in teaching profession since 2004. Presently I am working as a vocational subject teacher(IT) in a reputed govt. higher secondary school. Read more