```
#include <stdio.h>
#include <malloc.h>
void mult(int ,int* ,int );
int main()
{
int a,i,b[200],k,l,d,c,x,temp;
int *m;
scanf("%d",&a);
m = (int*) malloc (a*sizeof(int));
for(i=0;i<a;i++)
{
scanf("%d",m+i);
}
for(i=0;i<a;i++)
{
k=0;
l=*(m+i);
while(l/10!=0)
{
b[k]= l%10;
l=l/10;
k++;
}
b[k] = l;
for(d=*(m+i)-1;d>=1;d--)
{
temp=0;
for(c=0;c<=k;c++)
{
x=b[c]*d+temp;
b[c] = x%10;
temp = x/10;
}
while(temp!=0)
{
b[++k]=temp%10;
temp = temp/10;
}
}
for(c=k;c>=0;c--)
{
printf("%d",b[c]);
}
printf("\n",k);
}
return 0;
}
```

==============================================================================

```
#include <stdio.h>
#include <malloc.h>
void mult(int ,int* ,int );
int main()
{
int a,i,b[200],k,l;
int *m;
scanf("%d",&a);
m = (int*) malloc (a*sizeof(int));
for(i=0;i<a;i++)
{
scanf("%d",m+i);
}
for(i=0;i<a;i++)
{
k=0;
l=*(m+i);
while(l/10!=0)
{
b[k]= l%10;
l=l/10;
k++;
}
b[k] = l;
mult(*(m+i)-1,b,k);
}
return 0;
}
void mult(int i,int b[],int k)
{
int c,temp=0,l=0,x;
if(i!=1)
{
for(c=0;c<=k;c++)
{
x=b[c]*i+temp;
b[c] = x%10;
temp = x/10;
}
while(temp!=0)
{
b[++k]=temp%10;
temp = temp/10;
}
mult(i-1,b,k);
}
else
{
for(c=k;c>=0;c--)
{
printf("%d",b[c]);
}
printf("\n\n");
}
}
```

==============================================================================

In the above two codes the first code is executing perfectly but for the second one it is showing time limit exceeded. Why is that ? The logic I have used is same in both the cases.

asked
**02 May '12, 12:13**

0★Psyther

1●1●1●1

accept rate:
0%