大数的乘法

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 83811    Accepted Submission(s): 24704

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
原题：http://acm.hdu.edu.cn/showproblem.php?pid=1042
#include<stdio.h>#include<stdlib.h>#include<string.h>#include<iostream>using namespace std;#define max 10000#define base 100000//base取除10之外的值时要注意输出时除了第一个非0的数输出时不用注意格式//如果基数为10000，就要输出"%04d" ,因为会有前导0 ,基数越大计算次数越少 int num[max];int main(){int aim,i,j;while(scanf("%d",&aim)!=EOF){int carry=0;memset(num,0,sizeof(num));for(i=1,num[max-1]=1;i<=aim;i++){for(j=max-1;j>=0;j--){carry+=i*num[j];//不要忘了+  ，包含了上一位的进位 num[j]=carry%base;carry/=base;}}for(i=0;i<max && num[i]==0;i++); //找到第一个非0的数字，从此处开始输出 printf("%d",num[i++]);for(;i<max;i++)printf("%05d",num[i]);printf("\n");}return 0;}