ACM HDU 1042 N!（高精度计算阶乘）

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24579    Accepted Submission(s): 6774

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<stdio.h>
#include<string.h>
const int MAXN=40000;//如果是10000的阶乘，改为40000就够了 
int f[MAXN];
int main()
{
    int i,j,n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(f,0,sizeof(f));
        f[0]=1;
        for(i=2;i<=n;i++)
        {
            int c=0;
            for(j=0;j<MAXN;j++)
            {
                int s=f[j]*i+c;
                f[j]=s%10;
                c=s/10;
            }    
        } 
        for(j=MAXN-1;j>=0;j--) 
           if(f[j]) break;//忽略前导0
        for(i=j;i>=0;i--)  printf("%d",f[i]);
        printf("\n");   
    }    
    return 0;
}