洛谷 2257 - YY的GCD

莫比乌斯反演半模板题

很容易可以得到

\[Ans = \sum\limits_{p \in prime} \sum\limits_{d = 1}^{\min (\left\lfloor\frac{a}{p}\right\rfloor, \left\lfloor\frac{b}{p}\right\rfloor)} \mu(d) \left\lfloor\frac{a}{pd}\right\rfloor\left\lfloor\frac{b}{pd}\right\rfloor
\]

那么现在由于想要进行整除分块，所以希望将 \(\sum\) 内部的向下取整部分移到外部，故令 \(T = dp\) ，则有

\[\begin{aligned} Ans &= \sum\limits_{T = 1}^{\min (a, b)} \sum\limits_{p | T, p \in prime} \mu(\left\lfloor\frac{T}{p}\right\rfloor) \left\lfloor\frac{a}{T}\right\rfloor\left\lfloor\frac{b}{T}\right\rfloor \\ &= \sum\limits_{T = 1}^{\min (a, b)} \left\lfloor\frac{a}{T}\right\rfloor\left\lfloor\frac{b}{T}\right\rfloor \left( \sum\limits_{p | T, p \in prime} \mu(\left\lfloor\frac{T}{p}\right\rfloor) \right) \end{aligned}
\]

那么用筛法预处理一下 \(\mu\) 的那一部分就可以直接整除分块了

代码

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

typedef long long LL;

const int MAXN = 1e07 + 10;

int prime[MAXN];
int vis[MAXN]= {0};
int pcnt = 0;
int mu[MAXN]= {0};
LL tsum[MAXN]= {0}, sum[MAXN]= {0};
const int MAX = 1e07;
void prime_Acqu () {
	mu[1] = 1;
	for (int i = 2; i <= MAX; i ++) {
		if (! vis[i]) {
			prime[++ pcnt] = i;
			mu[i] = - 1;
		}
		for (int j = 1; j <= pcnt && i * prime[j] <= MAX; j ++) {
			vis[i * prime[j]] = 1;
			if (! (i % prime[j]))
				break;
			mu[i * prime[j]] = - mu[i];
		}
	}
	for (int j = 1; j <= pcnt; j ++)
		for (int i = 1; i * prime[j] <= MAX; i ++)
			tsum[i * prime[j]] += mu[i];
	for (int i = 1; i <= MAX; i ++)
		sum[i] = sum[i - 1] + tsum[i];
}

LL Calc (int a, int b) {
	LL ans = 0;
	int limit = min (a, b);
	for (int l = 1, r; l <= limit; l = r + 1) {
		r = min (a / (a / l), b / (b / l));
		ans += (sum[r] - sum[l - 1]) * (a / l) * (b / l);
	}
	return ans;
}

int T;

int getnum () {
	int num = 0;
	char ch = getchar ();

	while (! isdigit (ch))
		ch = getchar ();
	while (isdigit (ch))
		num = (num << 3) + (num << 1) + ch - '0', ch = getchar ();

	return num;
}

int main () {
	prime_Acqu ();
	T = getnum ();
	for (int Case = 1; Case <= T; Case ++) {
		int a = getnum (), b = getnum ();
		LL ans = Calc (a, b);
		printf ("%lld\n", ans);
	}

	return 0;
}

/*
2
10 10
100 100
*/