杭电 Problem 1024 N! 【大数相乘】

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 74388    Accepted Submission(s): 21600

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

 

Author

JGShining（极光炫影）

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <iostream>
#define MAX_N   100005
#define MAX(a, b) (a > b)? a: b
#define MIN(a, b) (a < b)? a: b
using namespace std;

int main() {
    int n;
    int m[MAX_N];
    while (scanf("%d", &n) != EOF) {
        m[0] = 1;
        int cnt = 1;
        for (int i = 1; i <= n; i++) {
            int q = 0;
            //用i乘以m[i]的每一位，不会超范围
            //并且为了保证m中的数均为一位数，要不断取余
            for (int j = 0; j < cnt; j++) {
                q = m[j]*i + q;
                m[j] = q%10;
                q /= 10;
            }
            //如果取余没有进行完，则剩下的数需要进位。
            while (q) {
                m[cnt++] = q%10;
                q /= 10;
            }
        }
        for (int i = cnt - 1; i >= 0; i--) {
            printf("%d", m[i]);
        }
        printf("
");
    }
    return 0;
}