[BZOJ2820]YY的GCD

题目大意

给你 \(n, m\) 求 $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^m [ gcd(x, y) == p ] $

解析

这道题实际上是上一道解决的HAOI2011 ProblemB的弱化版

同样的我们的我们可以设出 \(F(x)\) , \(f(x)\)，然后用莫比乌斯反演推出 \(f(x)\)

然后进行数论分块就行了

#include <bits/stdc++.h>
using namespace std;

int T;

const int maxn = 1e7 + 10;

int prime[maxn], mu[maxn], n, m, tot;
long long f[maxn];

inline void sieve() {
	mu[1] = 1; fill(prime, prime + maxn, 1); tot = 0;
	for (int i = 2; i < maxn; ++ i) {
		if (prime[i]) prime[++tot] = i, mu[i] = -1;
		for (int j = 1; j <= tot && i * prime[j] < maxn; ++ j) {
			prime[i * prime[j]] = 0;
			if (i % prime[j] == 0) {
				mu[i * prime[j]] = 0;
				break;
			} else {
				mu[i * prime[j]] = -mu[i];
			}
		}
	}

	for (int i = 1; i <= tot; ++ i) {
		for (int j = 1; j * prime[i] < maxn; ++ j) {
			f[j * prime[i]] += mu[j];
		}
	}
	for (int i = 1; i < maxn; ++ i) f[i] += f[i - 1];
}

int main() {
	sieve();
	scanf("%d", &T);
	while (T --) {
		scanf("%d%d", &n, &m);
		if (n > m) swap(n, m);
		long long ans = 0;
		for (int i = 1, nxt; i <= n; i = nxt + 1) {
			nxt = min(n / (n / i), m / (m / i));
			ans += (f[nxt] - f[i - 1]) * (n / i) * (m / i);
		}
		printf("%lld\n", ans);
	}
}