Answer / soumya

#include<stdio.h>
#include<conio.h>
void main()
{
int i,j;
clrscr();
for(i=100,j=1;i>0;i-=1,j+=1)
{
printf("%d,%d,",i,j);
}
getch();
}

Answer / vishwaranjan kumar singh

#include<stdio.h>
#include<conio.h>
void main()
{
int i=0,j=0;
getch();
for(i=100,j=1;i>0;i--,j++)
{
printf("%d-%d-",i,j);
}
clrscr();
}

Answer / niveditha

#include<stdio.h>
#include<conio.h>
void main()
{
int i,j,n,count=0;
printf("enter the number of no's to be displayed");
scanf("%d",&n);
for(i=100,j=1;count<n;i--,j++)
{
printf("%d,",i);
count=count+2;
if(count<=n)
printf("%d,",j);
}
getch();
}

Answer / suren

#include<stdio.h>
void main()
{
int i,j,n,count;
printf("Enter the number of numbers to be printed\n");
scanf("%d",&n);
for(i=100,j=1,count=0;count<n;i--,j++,count++)
{
printf(" %d ",i);
if(++count>=n)
return;
printf(" %d ",j);
}
}

Answer / gyanaranjan patra

#include<stdio.h>
void main()
{
static int a=100,i,b=1;
for (i=1;i<=100;i++)
{
printf("%d %d ",a,b);
a=a-1;
b=b+1;

}
}
// this is the simpler one try it;;;;.. ...

Answer / manohar bharti

#include<stdioh>
#include<conio.h>
void main()
{
int i,j,k,r;
printf("Enter the no. of terms you want");
scanf("%d",&r);
for(i=100,j=1,k=0;k<r;i--,j++,k++)
{
printf("%d",i);
k++;
printf(",");
if(k<r)
{
printf("%d",j);
printf(",");
}
else
return;
}
getch();
}

Answer / vishwaranjan singh

#include<stdio.h>
#include<conio.h>
void main()
{
int i=0,j=0;
getch();
for(i=100,j=1;i<0;i--,j++)
{
printf("%d-%d-",i,j);
}
clrscr();
}

int main() { int d = 10; int m = 2; int y = 3603; int c = 0; int val; val = ( d + m + y + (y/4) + c) % 7; cout << val; return 0; }

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