Problem b[HAOI2011]

题目描述

对于给出的 (n) 个询问，每次求有多少个数对 ((x,y))，满足 (a le x le b)，(c le y le d)，且 (gcd(x,y) = k)，(gcd(x,y)) 函数为 (x) 和 (y) 的最大公约数。

题解

莫比乌斯反演

我们可以用二维前缀和的思想，我们设

(f(n,m)=sumlimits_{i=1}^n sumlimits_{j=1}^m [gcd(i,j)=k])，

那答案应为

(f(b,d)-f(b,c-1)-f(a-1,d)+f(a-1,c-1))

接下来看看(f(n,m))怎么求：

(sumlimits_{i=1}^n sumlimits_{j=1}^m [gcd(i,j)=k])

(=sumlimits_{i=1}^{n/k} sumlimits_{j=1}^{m/k} [gcd(i,j)=1])

使用莫比乌斯反演

(=sumlimits_{i=1}^{n/k} sumlimits_{j=1}^{m/k} sumlimits_{d|gcd(i,j)}mu(d))

把(d)放到前面枚举，设(i=xd, j=yd)

(=sumlimits_{d} mu(d) * sumlimits_{x=1}^{n/kd} sumlimits_{y=1}^{m/kd} 1)

(=sumlimits_{d} mu(d) * lfloor frac{n}{kd} floor * lfloor frac{m}{kd} floor)

预处理(mu(d))的前缀和，使用除法分块即可做到时间复杂度(O(sqrt{n}))

总时间复杂度(O(nsqrt{n}))

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

int t, a, b, c, d, k;
int pr[50005], mb[50005], sum[50005], tot;
bool np[50005];

void init() {
	mb[1] = 1;
	for (int i = 2; i <= 50000; i++) {
		if (!np[i]) pr[++tot] = i, mb[i] = -1;
		for (int j = 1; j <= tot && i * pr[j] <= 50000; j++) {
			np[i*pr[j]] = 1;
			if (i % pr[j] == 0) {
				mb[i*pr[j]] = 0;
				break;
			} else mb[i*pr[j]] = -mb[i];
		}
	}
	for (int i = 1; i <= 50000; i++) sum[i] = sum[i-1] + mb[i];
}

int solve(int nn, int mm) {
	int ret = 0, n = nn / k, m = mm / k;
	for (int l = 1, r = 0; l <= min(n, m); l = r + 1) {
		r = min(n / (n / l), m / (m / l));
		ret += (sum[r] - sum[l-1]) * (n / l) * (m / l); 
	}
	return ret;
}

int main() {
	scanf("%d", &t);
	init();
	while (t--) {
		scanf("%d %d %d %d %d", &a, &b, &c, &d, &k);
		printf("%d
", solve(b, d) + solve(a-1, c-1) - solve(b, c-1) - solve(a-1, d));
	}
	return 0;
}