HDU 1042 N!(高精度计算阶乘)

• 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
  

  \

  题意：

  就是求n的阶乘。

  思路：

  开一个数组模拟乘法运算即可。

  #include<iostream>
  #include<cstdio>
  #include<string>
  #include<cmath>
  #include<cstring>
  using namespace std;
  const int mx = 41000;
  int a[mx];
  int main (){
  	int n;
  
  	while(scanf("%d",&n) != EOF){
  		memset(a, 0, sizeof(a));
  			a[0] = a[1]= 1;
  		for(int i=2 ;i <= n; i++){
  			int c = 0;
  			for(int j = 1; j < mx; j++){
  				int te = a[j] * i + c;
  				a[j] = te % 10;
  				c = te / 10;
  			}
  		}
  		int tt;
  		for(int i = mx-1; i>0; i--)
  			if(a[i] != 0) {
  				tt = i;
  				break;
  			} 
  		for(int i = tt; i > 0;i--)
  			printf("%d",a[i]);
  		puts("");
  		
  	}
  }
  

  
  \

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