POJ 3090 Visible Lattice Points (欧拉函数)

题目链接：POJ 3090

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 le x, y le 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 le x, y le N).

Input

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

Each dataset consists of a single line of input containing a single integer (N (1 le N le 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

Source

Greater New York 2006

Solution

题意

给定一个大小为 (N * N) 的矩形，每个格点插着钉子，问从 ((0, 0)) 点能看到多少钉子。

题解

欧拉函数

容易发现除了 ((0, 1)) ((1, 0)) ((1, 1))，如果一个钉子 ((x, y)) 能被看到，那么 (gcd(x, y) = 1, x eq y)。由于能看到的钉子关于过 ((0, 0)) 和 ((N, N)) 的直线对称，因此只考虑一半即可。对于每个 (x in [2, N])，需要求出多少个 (y) 满足 (gcd(x, y) = 1, 1 le y < x)，也就是求 (x) 的欧拉函数。

因此答案为 (3 + 2 * sum_{i=2}^N varphi(i))，其中 (varphi(i)) 为欧拉函数。

Code

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;

const int maxn = 1e4 + 10;

int primes[maxn], cnt;
int phi[maxn];
bool v[maxn];

void get_eulers(int n) {
    for(int i = 2; i <= n; ++i) {
        if(!v[i]) {
            primes[cnt++] = i;
            phi[i] = i - 1;
        }
        for(int j = 0; primes[j] <= n / i; ++j) {
            v[primes[j] * i] = 1;
            if(i % primes[j] == 0) {
                phi[primes[j] * i] = primes[j] * phi[i];
                break;
            }
            phi[primes[j] * i] = phi[i] * (primes[j] - 1);
        }
    }

}

int main() {
    get_eulers(1001);
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    int kase = 0;
    while(T--) {
        int n;
        cin >> n;
        ll sum = 0;
        for(int i = 2; i <= n; ++i) {
            sum += phi[i];
        }
        cout << ++kase << " " << n << " ";
        cout << 3 + 2 * sum << endl;
    }
    return 0;
}