N!---hdu-1042

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 65262    Accepted Submission(s): 18665

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

 

#include<stdio.h>
#include<string.h>
#define max 36000
int main()
{
    int a[max],i,j,n,k,p;
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        a[1]=1;
        int c;
        for(i=2;i<=n;i++)
        {
            c = 0;
            for(j=1;j<max;j++)
            {
                a[j] = a[j] * i + c;
                c = a[j] / 10;
                a[j] = a[j] % 10;
            }
        }
        for(i=max-1;(a[i]==0)&&(i>=0);i--); 
        if(i>=0)
        {
            for(;i>=1;i--)
            printf("%d",a[i]);
        }
        printf("\n");
        }
    return 0;
}

这题的思路就是2*3*4*5*...n。把所乘得的数存放在数组中，记得进位！！！

再放个我最开始的代码，哎，超时的伤，你不懂！

#include<stdio.h>
#include<string.h>
#define max 36000
int main()
{
    int a[max],i,j,n,k,p;
    while(scanf("%d",&n),n>=0)
    {

        memset(a,0,sizeof(a));
        a[1]=1;
        for(i=2;i<=n;i++)
        {
            for(k=1;k<max;k++)
            a[k]*=i;
            for(j=1;j<max;j++)
            {
                if(a[j]>=10)
                {
                a[j+1]+=a[j]/10;
                a[j]%=10;    
                }
            }
        }
        for(i=max-1;(a[i]==0)&&(i>=0);i--); 
        if(i>=0)
        {
            for(;i>=1;i--)
            printf("%d",a[i]);
        }
        printf("\n");
    
    }
    return 0;
}