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generate this output using for loop
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Answer / varsha shukla
#include<conio.h>
#include<stdio.h>
void main()
{
int i,j,k,l;
for(i=1;i<=4;i++)
{
for(k=i;k<=4;k++)
{
printf(" ");
}
for(j=1;j<=i;j++)
{
printf(" %d",j);
}
printf("\n");
}
for(i=3;i>0;i--)
{
for(l=i;l<=3;l++)
{
printf(" ");
}
for(j=1;j<=i;j++)
{
printf(" %d",j);
}
printf("\n");
}
getch();
| Is This Answer Correct ? | 4 Yes | 1 No |
Answer / narasimharao
#include<stdio.h>
#include<conio.h>
void main()
{
int i,j,k,n;
printf("Enter n value:");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(k=i;k<n;k++)
{
printf(" ");
}
for(j=1;j<=i;j++)
{
printf("%d",j);
for(k=i;k<i+1;k++)
{
printf(" ");
}
}
printf("\n");
}
for(i=n-1;i>=1;i--)
{
for(k=i;k<n;k++)
{
printf(" ");
}
for(j=1;j<=i;j++)
{
printf("%d",j);
for(k=i;k<i+1;k++)
{
printf(" ");
}
}
printf("\n");
}
getch();
}
| Is This Answer Correct ? | 1 Yes | 2 No |
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