扩展1000!(n!)的尾数零的个数

#include <stdio.h>
#include <malloc.h>
//计算1000!尾数零的个数
//扩展n!的尾数零的个数
//2^a * 5^b
//obviously a>b
//so count = b
//5*1 5*2 .....5*200  200个  除以5
//1 2 3 4 5 .... 200 中 包含 5*(1,2,3,4,...40) 40个 除以5
//5*1 5*2 ..... 5*8

void calc1(int n){
    int count=;
    int i=,t;
    for(;i<=n;i++)
    {
        t = i;
        while(t%==)
        {
            count++;
            t = t/;
        }
    }
    printf("%d",count);
}

void calc2(int n)
{
    int sum=;
    while(n)
    {
        n/=;
        sum+=n;
    }
    printf("%d",sum);
}

int main () {
    int n = ;
    calc2(n);
    return ;
}