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# Program for trailing zeroes

Andrew Musgrave
Greenhorn
Posts: 8
#include<math.h>
#include<stdio.h>
int main()
{
long i,j[100000],k,l,T,a,n[100000],c,sum=0,m;
scanf("%ld\n", &T);
for(c=1; c<=T; c++)
{
n[c] = 0;
}
for(i=1; i<=T; i++)
{
scanf("%ld", &j[i]);
printf(" \n");
}
for(k=1; k<=T ; k++)
{
for(l=1; j[k]>pow(5,l); l=l+15);
a=l++;
n[k] = j[k];
while(a>0)
{
n[k] = n[k]/pow(5,l);
sum = sum + n[k];
}
n[k] = sum;
}
for(m=1; m<=T; m++)
{
printf("%ld \n", &n[m]);
}
return 0;
}
long pow(long x,long y)
{
if(y==0)
return 1;
else
return(x*pow(x,y-1));
}

Error :Conflicting types for POW

And is the program correct otherwise?
Here is the program statement
Input

There is a single positive integer T on the first line of input (equal to about 100000). It stands for the number of numbers to follow. Then there are T lines, each containing exactly one positive integer number N, 1 <= N <= 1000000000.

Output

For every number N, output a single line containing the single non-negative integer Z(N).
Example

Sample Input:

6
3
60
100
1024
23456
8735373

Sample Output:

0
14
24
253
5861
2183837

Andrew Musgrave
Greenhorn
Posts: 8
Please Reply. I know, we have to be patient, but this problem has me bugged since last week.

Anand Hariharan
Rancher
Posts: 272
Andrew Musgrave wrote:Please Reply. I know, we have to be patient, but this problem has me bugged since last week.

You could help yourself by doing the following; you are likely to receive some response that way:

• Format your code so as to be readable. There is a feature called CodeTags that should help.
• Explain what Z(n) is. Your subject hints at something about trailing zeros, but that doesn't help either.

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