Poj3090 Visible Lattice Points 欧拉函数求和

最新推荐文章于 2021-05-10 16:41:07 发布

Megumin

最新推荐文章于 2021-05-10 16:41:07 发布

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本文介绍了一种计算在第一象限中从原点可见的格点数量的方法，并提供了一个具体的程序实现，该程序能有效计算指定范围内不被其他格点遮挡的点的数量。

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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.

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

4
2
4
5
231

Sample Output

1 2 5
2 4 13
3 5 21
4 231 32549

求两点连线不经过其他整点的点的个数，即求欧拉函数和。

#include <iostream>
#include <cstdio>
#include <map>
#include <cmath>
#include <map>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
#define LL long long
#define maxn 1009
int euler[maxn],ans[maxn];
void euler_init(){
    int i,j;
    euler[1]=1;
    for(i=2;i<maxn;i++)
      euler[i]=i;
    for(i=2;i<maxn;i++)
       if(euler[i]==i)
          for(j=i;j<maxn;j+=i)
            euler[j]=euler[j]/i*(i-1);
    memset(ans,0,sizeof(ans));
    for(int i=0;i<=1000;i++){
        for(int j=0;j<=i;j++){
            ans[i]+=euler[j];
        }
        ans[i]=2*ans[i]+1;
    }
}
int main()
{
    int t,n,ca=1;
    euler_init();
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        printf("%d %d %d\n",ca++,n,ans[n]);
    }
    return 0;
}