题目大意

N!末尾0的个数

题解

0只能由2*5产生,所以只要求2,5有多少对即可,又因为10!中5的个数少于2,所以只要求因子5有多少个即可,答案即为N/5+N/25+N/125..

代码:

#include<stdio.h>
int main(void)
{
int T;
scanf("%d",&T);
while(T--)
{
int n,ans=0;
scanf("%d",&n);
while(n)
{
ans+=n/5;
n/=5;
}
printf("%d\n",ans);
}
return 0;
}

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