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 ;

}