• 博客园logo
  • 会员
  • 众包
  • 新闻
  • 博问
  • 闪存
  • 赞助商
  • HarmonyOS
  • Chat2DB
    • 搜索
      所有博客
    • 搜索
      当前博客
  • 写随笔 我的博客 短消息 简洁模式
    用户头像
    我的博客 我的园子 账号设置 会员中心 简洁模式 ... 退出登录
    注册 登录
james1207

博客园    首页    新随笔    联系   管理    订阅  订阅

11417 - GCD

Problem A
GCD
Input: 
Standard Input

Output: Standard Output

 

Given the value of N, you will have to find the value of G. The definition of G is given below:

 

Here GCD(i,j) means the greatest common divisor of integer i and integer j.

 

For those who have trouble understanding summation notation, the meaning of G is given in the following code:

G=0;

for(i=1;i<N;i++)

for(j=i+1;j<=N;j++)

{

    G+=GCD(i,j);

}

/*Here GCD() is a function that finds the greatest common divisor of the two input numbers*/

 

Input

The input file contains at most 100 lines of inputs. Each line contains an integer N (1<N<501). The meaning of N is given in the problem statement. Input is terminated by a line containing a single zero.  This zero should not be processed.

 

Output

For each line of input produce one line of output. This line contains the value of G for corresponding N.

 

Sample Input                              Output for Sample Input

10

100

500

0

 

67

13015

442011

#include<stdio.h>
int gcd(int a, int b)
{
	if(!b) return a;
	return gcd(b,a%b);
}
int main()
{
	int n;
	while (scanf("%d",&n)&&n)
	{
		int g=0,i,j;
		for(i=1;i<n;i++)
			for(j=i+1;j<=n;j++)
				g+=gcd(j,i);
		printf("%d\n",g);
	}
	return 0;
}
posted @ 2013-09-15 19:28  Class Xman  阅读(237)  评论(0)    收藏  举报
刷新页面返回顶部
博客园  ©  2004-2025
浙公网安备 33010602011771号 浙ICP备2021040463号-3