N!
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53785 Accepted Submission(s): 15217

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
  
    
#include
   
     const int maxn=50000; //数组开到50000就可以满足10000的阶乘不越界 int fun[maxn]; int main() { int i,j,n; while(~scanf("%d",&n)) { memset(fun,0,sizeof(fun)); fun[0]=1; for(i=2;i<=n;i++) //从2的阶乘开始，一直到指定数的阶乘 { int c=0; for(j=0;j
    
     =0;j--) //找出该数的最高位，即数组角码最大且不为0的数 if(fun[j]) break; for(i=j;i>=0;i--) printf("%d",fun[i]); printf("\n"); } return 0; }