N!

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 83996 Accepted Submission(s): 24766

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）

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分析：一开始以为要用高精度，然后看了下其他人的做法，原来可以直接用数组储存当前位

然后又因为不知道 0!=1 WA了很多发..

注意，一开始数组需要多长的话，可能需要试探性的取吧，再看ans[MAXN]是否为0

代码如下：

#include 
      
       #include 
       
        #include 
        
         #include 
         #include 
          
           using namespace std;typedef long long ll;const int MAXN=40000;int ans[MAXN];int flag;int main(){    int n,r,tmp;while(scanf("%d",&n)!=EOF){    flag=0;     memset(ans,0,sizeof(ans));     ans[0]=1;     for(int i=2;i<=n;i++)     {         r=0;        for(int j=0;j<=MAXN;j++)        {            tmp=ans[j]*i+r;            ans[j]=tmp%10;            r=tmp/10;        }     }       for(int i=MAXN;i>=0;i--)       {          if(flag==0&&ans[i]==0)          continue;          else          {              flag=1;              printf("%d",ans[i]);          }       }       printf("\n");    }    return 0;}