POJ3090 Visible Lattice Points （数论：欧拉函数模板）

题目链接：传送门

思路：

　　所有gcd(x, y) = 1的数对都满足题意，然后还有(1, 0) 和 (0, 1)。

#include <iostream>
#include <cstring>

using namespace std;
const int MAX_N = 1e3 + 5;
int prime[MAX_N+1], phi[MAX_N+1];
void getPrime_and_Phi() {
    memset(prime, 0, sizeof prime);
    phi[1] = 1;
    for (int i = 2; i <= MAX_N; i++) {
        if (!prime[i]) prime[++prime[0]] = i, phi[i] = i-1;
        for (int j = 1; j <= prime[0] && prime[j] <= MAX_N/i; j++) {
            prime[i*prime[j]] = 1;
            phi[i*prime[j]] = phi[i] * (i%prime[j] ? prime[j]-1 : prime[j]);
            if (i%prime[j] == 0) break;
        }
    }
}

int main()
{
    getPrime_and_Phi();
    int C;
    cin >> C;
    for (int kase = 1; kase <= C; kase++) {
        int N;
        cin >> N;
        long long ans = 1;
        for (int i = 1; i <= N; i++)
            ans += phi[i]*2;
        cout << kase << ' ' << N << ' ' << ans << endl;
    }
    return 0;
}

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