P8 Visible Lattice Points

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原文链接：http://www.cnblogs.com/yaobolove/p/4155170.html

本文介绍了一种计算第一象限内可见格点数量的算法。对于大小为N的区域，通过判断各点坐标是否互质来确定可见性，并统计可见点总数。提供了一个具体的实现代码示例。

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P8 Visible Lattice Points

Time Limit:1000ms, Memory Limit:65536KB

Description

A lattice point (x, y) in the first quadrant (x and y are integers greater than or equal to 0), other than the origin, is visible from the origin if the line from (0, 0) to (x, y) does not pass through any other lattice point. For example, the point (4, 2) is not visible since the line from the origin passes through (2, 1). The figure below shows the points (x, y) with 0 ≤ x, y ≤ 5 with lines from the origin to the visible points.

图片链接：http://blog.sina.com.cn/s/blog_4a7304560101ajjf.html

Write a program which, given a value for the size, N, computes the number of visible points (x, y) with 0 ≤ x, y ≤ N.

Input

The first line of input contains a single integer C (1 ≤ C ≤ 1000) which is the number of datasets that follow.

Each dataset consists of a single line of input containing a single integer N (1 ≤ N ≤ 1000), which is the size.

Output

For each dataset, there is to be one line of output consisting of: the dataset number starting at 1, a single space, the size, a single space and the number of visible points for that size.

Sample Input

231

Sample Output

1 2 5

2 4 13

3 231 32549

分析：

主要求法：是用两个互质因数的求解，我个人认为有求最大公约数的方法进行求解！！！

#include<iostream>
#include<math.h>
using namespace std;
int main()
{
    int n,m,k=0;
    int x,y;
    int count;
    int prime(int,int );
cin>>m;
    while(m--)
    {
        cin>>n;
        count=2;
      k++;
    if(n==1)
    cout<<k<<" "<<n<<" "<<"3";


    else if(n>1)
    {

        for(x=1;x<=n;x++)
        {

        for(y=1;y<=n;y++)
        {

          if(prime(x,y)==1)
             count++;
        }

        }


        cout<<k<<" "<< n<<" "<<count;
       // cout<<endl;

    }

    }
 return 0;


}
int prime(int u,int v)
{
int t,r;
r=1;
if(v>u)
{
    t=u;u=v;v=t;}
    while((r=u%v)!=0)
    {
        u=v;
        v=r;
    }
  return v;
    }

转载于:https://www.cnblogs.com/yaobolove/p/4155170.html