3090 Visible Lattice Points 一个(n+1)*(n+1)的点阵，问多少点能被点(0,0)看到 欧拉函数

Visible Lattice Points

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3212		Accepted: 1761

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

 

题目大意：一个(n+1)*(n+1)的点阵，问多少点能被点(0,0)看到。如果(0,0)到(i,j)的连线被点挡住就算看不到。

先画一条(0, 0)到(n, n)的线，把图分成两部分，两部分是对称的，只需算一部分就好。取右下半，这一半里的点(x, y)满足x >= y可以通过欧拉函数计算第k列有多少点能够连到(0, 0)若x与k的最大公约数d > 1，则(0, 0)与(x, k)点的脸先必定会通过(x/d, k/d)，就被挡住了所以能连的线的数目就是比k小的、和k互质的数的个数，然后就是欧拉函数。由于是对称的，所以只需算一半的数量就够了。第k列上的一个点的纵坐标为d,若gcd(k,d) != 1,则远带你与该点的连线必须通过（k/gcd,d/gcd）,肯定被挡住了。

由此我们可以得到递推公式 res1[i] = res1[i-1] + 2*phi[i];其中phi[i]是第i列上能看到的点的个数，

这是欧拉函数的解法：

 

 

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn = 1005;
int phi[1005];
int a[1005];
void euler()
{
    int i,j;
    for(i=1; i<=maxn; i++)
        phi[i] = i;
    for(i=2; i<=maxn; i+=2)
        phi[i] /= 2;
    for(i=3; i<=maxn; i+=2)
        if(phi[i] == i)
        {
            for(j=i; j<=maxn; j+=i)
                phi[j] = phi[j] / i * (i - 1);
        }
}
int main()
{
    int t,n,i,j;
    euler();
    a[1] = 3;
    for(i=2; i<=1000; i++)
        a[i] = a[i-1] + phi[i]*2;
    scanf("%d",&t);
    for(i=1; i<=t; i++)
    {
        scanf("%d",&n);
        printf("%d %d %d/n",i,n,a[n]);
    }
    return 0;
}