Problem Description

Givenan integer N(0 ≤ N ≤ 10000), your task is to calculate N!

Input

OneN in one line, process to the end of file.

Output

Foreach N, output N! in one line.

Sample Input

1

2

3

Sample Output

1

2

6

Author

JGShining（极光炫影）

/*********

高精度数，大数阶乘

类比十进制，模拟一个万进制的算法算法，参考：http://blog.csdn.net/lulipeng_cpp/article/details/7437641

用int 存数位，10000*100000<2^31，所以我每一位存了 100000，当然与 10000是类似的

*********/

Code:

#include<iostream>

#include<stdio.h>

#include<string.h>

#include<cmath>

using namespace std;

/**

计算阶乘位数，顺便把POJ 1423 给A了

#define PI 3.141592653589793239

#define ee 2.7182818284590452354

int ans(int n){

    return (int)((n*log10(n/ee)+log10(sqrt(2*n*PI))))+1;

}

*/

int num[8000];

int main()

{

    int n,i,j;

    while(cin>>n)

    {

        memset(num,0,sizeof(num));

        num[0] = 1;

        for(i = 2;i<=n;i++)

        {

            for(j = 0;j<8000;j++)

                num[j]*=i;

            for(j = 0;j<8000;j++)

            {

                num[j+1] += num[j]/100000;

                num[j] %= 100000;

            }

        }

        int len = 8000;

        while(num[len] == 0)

        {

            len--;

        }

        cout<<num[len];

        for(i = len-1;i>=0;i--)

        {

            printf("%.5d",num[i]);

        }

        printf("\n");

    }

    return 0;

}