HDU1042 N!

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 82480    Accepted Submission(s): 24200

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

如果正常的的话，递归或者迭代

// 递归计算阶乘
int factorial_recursion(int n){
    if(n<=0){
        return 1;
    }else{
        return n * factorial_recursion(n-1);
    }
}
// 迭代计算阶乘
int factorial_iteration(int n){
    int result = 1;
    while(n>1){
        result *= n;
        n--;
    }
    return result;
}

不过int不超过14就GG了。

所以还是用数组去模拟，用数组模拟了竖式除法，再输出结果

/*************************************************************************
    > File Name: N的阶乘.c
    > Author: YinJianxiang
    > Mail: YinJianxiang123@gmail.com 
    > Created Time: 2017年06月20日 星期二 21时54分10秒
 ************************************************************************/

#include<stdio.h>
#include<string.h>

int main(void) {
    long long result[10000];
    int num;
    int i;
    int j;
    while(scanf("%d",&num) != EOF) {
        memset(result,0,sizeof(result));
        result[0] = 1;
        for(j = 1;j < num;j++) {
            for(i = 0;i < 9999;i++) {
                result[i] *= j;
                for(i = 0;i < 9999;i++) {
                    if(result[i] >= 10000) {
                        result[i+1] += result[i] / 10000;
                        result[i] %= 10000;
                    }
                }
            }
        }
 
        for(i = 9999;i >= 0;i--) {
            if(result[i]) {
                break;
            }
        }
        printf("%lld",result[i]);
        for(j = i - 1;j >= 0;j--) {
            printf("%05lld",result[i]);
        }
        printf("\n");
    }
    
    return 0;
}