N!

题目来自杭电：http://acm.hdu.edu.cn/showproblem.php?pid=1042
Problem 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

代码：

#include <iostream>
#include <cstring>
#include <fstream>
using namespace std;

int main()
{
    int n;
    int i;
    int j;
    int digit;  //总位数
    int a[40000];
    int carry;  //进位
    int temp;
    int len;
    //ofstream cout("out.txt");
    while(cin >> n){
        memset(a,0,sizeof(a));
        temp = 0;
        digit = 1;
        a[0] = 1;
        for(i = 2;i <= n;i++){
            carry = 0;
            for(j = 1;j <= digit;j++){
                temp = a[j-1] * i + carry;
                a[j-1] = temp % 10;
                carry  = temp /10;
            }
            while(carry){
                a[++digit - 1] = carry % 10;
                carry /= 10;
            }
        }
        for(i = digit-1;i >= 0;i--)
            cout << a[i];
        cout << endl;

    }
    return 0;
}

总结：还是大数问题，一般情况下12的阶乘就已经溢出，因此考虑用数组存储，关于大数阶乘，有很多很多好的算法，此处只是个很简单的，仅仅比递归想的多了点，奉上大神之作http://blog.youkuaiyun.com/whdugh/article/details/9364245