HDoj-1042 大數階乘

標籤：c語言   algorithm   namespace   printf   c   

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<stdio.h> #include<string.h>const int maxn=50000;    //數組開到50000就可以滿足10000的階乘不越界 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++)          //從2的階乘開始，一直到指定數的階乘    {   int c=0;    for(j=0;j<maxn;j++)  //將所得階乘數放在fun數組中，低位放在fun[0]中     {    int s=fun[j]*i+c;      fun[j] =s%10;      c=s/10;        }       }         for(j=maxn-1;j>=0;j--)     //找出該數的最高位，即數組角碼最大且不為0的數             if(fun[j])  break;  for(i=j;i>=0;i--)            printf("%d",fun[i]);   printf("\n");}return 0;}

HDoj-1042 大數階乘