HDU 1042 N！ N的阶乘大数

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 75303    Accepted Submission(s): 21981

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

求n的阶乘嘛！ 基础题，依然WA，也是够了！

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#include<queue>
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f

#define N 1100000
int a[N];

int main()
{
    int i,j,n,count,t,k;
    while(scanf("%d",&n)!=EOF)
    {
        a[0]=1;
        count=1;
        for(i=1;i<=n;i++)
        {
            k=0;
            for(j=0;j<count;j++)
            {
                t=a[j]*i+k;
                a[j]=t%10;
                k=t/10;
            }
            while(k)
            {
                a[count++]=k%10;
                k=k/10;
            }
        }
        for(i=count-1;i>=0;i--)
        {
            printf("%d",a[i]);
        }
        printf("\n");
    }
    return 0;
}

10000的阶乘也就35660位，要是担心TLE，或者超限的话，也可以按10000进制来写：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#include<queue>
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f

#define N 1100000
int a[N];

int main()
{
    int i,&n)!=EOF)
    {
        a[0]=1;
        count=1;
        for(i=1;i<=n;i++)
        {
            k=0;
            for(j=0;j<count;j++)
            {
                t=a[j]*i+k;
                a[j]=t%10000;
                k=t/10000;
            }
            while(k)
            {
                a[count++]=k%10000;  ///这里以10000为进位单位
                k=k/10000;
            }
        }
        printf("%d",a[count-1]);  ///注意一下，第一个数不是“%04d”；
        for(i=count-2;i>=0;i--)
        {
            printf("%04d",a[i]);  ///其他的按四位的输出，不足补零
        }
        printf("\n");
    }
    return 0;
}