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

123

 

Sample Output

126

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: [e-mail]     > 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;}