Class 10 Computer Science Chapter 4 | NESTED LOOPS IN C: Welcome to our website! We are happy to provide you with Class 10th Notes for your academic journey. Class IX is an essential year for learners as they prepare themselves for the upcoming board exams, which will determine their academic success.
Today, I will discuss your Class 10th “NESTED LOOPS IN C” Long and Short Questions| We provide solutions for almost all long and short questions.
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TEXTUAL QUESTIONS AND ANSWERS
EXERCISE
1. Write C programs to display the following pattern using nested loop construct.
(a)
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
Solution:
#include<stdio.h>
int main()
int i, j;
for(i=1;i<=5; i++)
{
for(j=1; j<=3;j++)
{
printf(“%d “j);
}
printf(“\n”);
}
return 0;
Output:
1 2 1
1 2 1
1 2 1
1 2 1
1 2 1
Process returned 0 (0x0) execution time: 0.311 s
Press any key to continue.
(c) 4 3 2 1
4 3 2 1
4 3 2 1
4 3 2 1
4 3 2 1
Solution:
#include<stdio.h>
int main()
{
int i, j;
for(i=1; i<=5; i++)
{
for(j=4; j>=1; j–)
{
printf(” %d “,j);
}
printf(“\n”);
}
return(0);
Output:
4 3 2 1
4 3 2 1
4 3 2 1
4 3 2 1
4 3 2 1
Process returned 0 (0x0) execution time: 0.638 s
Press any key to continue.
(d)
2
2 3 4
2 3 4 5 6
Solution:
Code:
#include<stdio.h>
int main()
{
int i, j, k, m=2;
for(i=3; i>=1; i–)
{
for(j=1; j<I; j++)
{
printf(“j”);
}
for(k=2;k<=m;k++)
{
printf(“%d”,k);
}
m = m + 2;
printf(“\n”);
}
return(0);
}
Output:
2
2 3 4
2 3 4 5 6
Process returned 0 (0x0) execution time: 0.504 s
Press any key to continue.
(e)
1
1 2 1
1 2 3 2 1
Solution:
Code:
#include<stdio.h>
int main()
{
int i, j, k, p;
for(k=1;k<=3;k++)
{
for(i=1;i<=3-k;i++)
{
printf(” “);
}
for(j=1;j<=k;j++)
{
printf(” %d “, j);
}
for(p=1; p<=k-1; p++)
{
printf(” %d “,p);
}
printf(“\n”);
}
return 0;
}
Output:
1
1 2 1
1 2 3 1 2
Process returned 0 (0x0) execution time: 0.304 5
Press any key to continue.
(f)
.
. . .
. . . . .
. . . . . . .
. . . . .
. . .
.
Solution:
Code:
#include<stdio.h>
int main()
{
int i, j, k, p;
for(k=1;k<=4;k++)
for(i=1;i<=4-k;i++)
{
printf(” “);
}
{
for(j=1;j<=k;j++)
{
printf(“* “);
}
for(p=1;p<=k-1;p++)
{
printf(” “);
}
printf(“\n”);
}
for(k=3;k>=1;k-)
{
for(i=1;i<=4-k; i++)
{
printf(” “);
}
for(j=1;j<=k;j++)
{
printf(“* “);
for(p=1;p<=k-1;p++)
{
printf(“* “);
}
printf(“\n”);
}
return 0;
}
Output:
.
. . .
. . . . .
. . . . . . .
. . . . .
. . .
.
Process returned 0 (0x0) execution time: 0.620 s
Press any key to continue.
(g)
x
x x
x x x
x x x x
x x x x x
x
x x
x x x
x x x x
x x x x x
Solution:
Code:
#include<stdio.h>
int main()
{
int i, j, k, p=4;
for(k=1;k<=5;k++)
for(i=1;i<=k; i++)
{
printf(” X “);
}
for(j=1; j<=p; j++)
{
printf(” “);
}
for(i=1;i<=k; i++)
{
printf(” X “);
}
printf(“\n”);
}
return 0;
Output:
{
Process returned 0 (exe) execution time: 0.402 s
Press any key to continue.
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.
Solution:
(a) Code:
#include<stdio.h>
int main()
{
int I,j,n;
print(“Enter the number of lines: “);
scanf(“%d”,&n);
for(i=1;i<=n; i++)
{
for(j=1;j<=3;j++)
}
printf(“%d “,j);
}
printf(“\n”);
}
return(0);
}
}
Output:
Enter the number of lines : 8
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
Process returned 0 (exe) execution time : 4.187 s
Press any key to continue.
(b) Code:
#include<stdio.h>
int main()
{
int i, j, n;
printf(“Enter the number of lines:”);
scanf(“%d”, &n);
for(i=1;i<=n; i++)
{
for(j=1; j<=2;j++)
{
printf(“%d “,j);
}
for(j=1;j>=1;j-)
{
printf(“ %d”,j);
}
printf(“\n”);
}
return(0);
Conclusion:
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