Description:

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

Input:

One N in one line, process to the end of file.

Output:

For each N, output N! in one line.

Sample Input:

1
2
3

Sample Output:

1
2
6

题目链接

数的阶乘计算起来可能会很大，通常会因为爆数据范围无法存储，万进制求阶乘就是用一个结果数组存阶乘结果，数组中每一个元素只记录结果的四位数，这样结果就可以很好的保存了。
思路:对n从1开始求阶乘，如果结果大于10000则将大于的部分在前一个res数组元素中保存，以此类推用
res计算阶乘结果。输出的时候最左边的四个数单独输出，后面的每个4位数如果前几位为0则需要在前面补零补齐4位输出。

AC代码:

#include <stdio.h>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iomanip>
#include <cctype>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdlib>

using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long ll;

int res[10010];
int n;

void Solve() {
    int Book = 1;
    int BaoFour = 0;
    res[Book] = 1;
    for (int i = 1;i <= n;++i) {
        BaoFour = 0;
        for (int j = 1;j <= Book;++j) {
            res[j] = res[j] * i + BaoFour;
            BaoFour = res[j] / 10000;
            res[j] = res[j] % 10000;
        }
        if (BaoFour > 0) {
            res[++Book] += BaoFour;
        }
    }
    cout << res[Book];
    for (int i = Book - 1;i > 0;--i) {
        if (res[i] >= 1000) {
            cout << res[i];
        }
        else if (res[i] >= 100) {
            printf("0%d",res[i]);
        }
        else if (res[i] >= 10) {
            printf("00%d",res[i]);
        }
        else {
            printf("000%d",res[i]);
        }
    }
    cout << endl;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    while (cin >> n) {
        mem(res, 0);
        Solve();
    }
    return 0;
}