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算法思想：在1-10两个数相乘要产生0，只有 10×1=2×5，2×5。
200！=200×199×198……×2×1=2×5×2×5×2×199…. ×2×1;可以分解为质数相乘的形式,
很明显有2的个数比5的多,所以只要求出200的阶乘可分解出多少个5相乘,
就可得到200的阶乘结尾的连续的零的个数.
即:num=[200/5]+[200/5/5]+[200/5/5/5].
注: [x]表示对x取整.
所以可以通过这个思路很容易的得到任意阶乘结尾连续的零，其示例C语言代码如下：
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#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
using namespace std;
#define MAX 100
//MAX 是计算结果的最大位数 取为100位
void factorial(int n,char output[])
{
int i, j, cin, tmp;
int result[MAX];
memset(result, 0, sizeof(result)); //初始化为0
result[0] = 1;
for(i = 2; i <= n; ++i) //从2 开始计算阶乘
{
cin = 0; //进位初始化为0
for(j = 0; j < MAX; ++j)
{
tmp = result[j] * i + cin; //cin进位
result[j] = tmp % 10;
cin = tmp / 10;
}
}
for(i = MAX - 1; i >= 0; --i) //忽略前导的0
if(result[i]) break;
for(j = i; j >= 0; j--) //将结果倒叙打印在结果数组中
sprintf(output++,"%d", result[j]);
//注意sprintf的第一个参数是char型指针
}
/*计算n!结尾零的个数，返回结尾零的个数。*/
int GetZeroNum(int n)
{
int num=0;//n!结尾零的个数
int b=1;//5的次方
while(1)
{
b*=5;
num+=n/b;
if(b>n)
break;
}
return num;//返回结尾零的个数
}
int main()
{ char out[MAX]={0};
int i=5;
factorial(i,out);
cout<<"factorial("<<i<<") is "<<out<<endl;
cout<<"factorial("<<i<<") the number of 0 in the rear is "<<GetZeroNum(i)<<endl;
i=10;
factorial(i,out);
cout<<"factorial("<<i<<") is "<<out<<endl;
cout<<"factorial("<<i<<") the number of 0 in the rear is "<<GetZeroNum(i)<<endl;
i=20;
factorial(i,out);
cout<<"factorial("<<i<<") is "<<out<<endl;
cout<<"factorial("<<i<<") the number of 0 in the rear is "<<GetZeroNum(i)<<endl;
i=30;
factorial(i,out);
cout<<"factorial("<<i<<") is "<<out<<endl;
cout<<"factorial("<<i<<") the number of 0 in the rear is "<<GetZeroNum(i)<<endl;
}
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factorial(5) is 120
factorial(5) the number of 0 in the rear is 1
factorial(10) is 3628800
factorial(10) the number of 0 in the rear is 2
factorial(20) is 2432902008176640000
factorial(20) the number of 0 in the rear is 4
factorial(30) is 265252859812191058636308480000000
factorial(30) the number of 0 in the rear is 7
Process returned 0 (0x0) execution time : 0.109 s
Press any key to continue.
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