大数阶乘位数

发布时间：2020-12-14 03:16:45 所属栏目：大数据来源：网络整理

导读：N! (N factorial) can be quite irritating and difficult to compute for large values of N. So instead of calculating N!,I want to know how many digits are in it. (Remember that N! = N * (N - 1) * (N - 2) * ... * 2 * 1) Input Each line of the

N! (N factorial) can be quite irritating and difficult to compute for large values of N. So instead of calculating N!,I want to know how many digits are in it. (Remember that N! = N * (N - 1) * (N - 2) * ... * 2 * 1)

Input

Each line of the input will have a single integer N on it 0 < N < 1000000 (1 million). Input is terminated by end of file.

Output

For each value of N,print out how many digits are in N!.

Sample Input

1
3 
32000

Sample Output

1 
1 
130271 

原理：利用迭代，原本计算公式应该是(n*(n-1)*...*2*1)/(10*10*...*10)，计算能除多少个10，结果再加1.
现在是从第一个n开始就除10，判断能否除完以后的结果>1,如果大于1，就result++。

代码：
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <stack>
#include<cmath>
using namespace std;
int main()  
{  
    int n ;  
    while(scanf("%d",&n)!=EOF)  
    {  
        int result=1;  
        int i;  
        double temp = 1;  
        for(i=n;i>0;i--)  
        {  
            temp*=i;  
            while((temp/10)>1)  
            {  
                result++;  
                temp=temp/10;  
            }  
        }  
        printf("%dn",result);  
    }  
    return 0;  
}

（编辑：李大同）

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