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.csdn.net/whdugh/article/details/9364245