题目

P2522 [HAOI2011]Problem b

解析：

具体推导过程同P3455 [POI2007]ZAP-Queries

不同的是，这个题求的是\(\sum_{i=a}^b\sum_{j=c}^dgcd(i,j)=k\)

像二维前缀和一样容斥一下，输出就完了。

根据luogu某大佬的说法



开longlong的话会TLE。。

代码

//莫比乌斯反演

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 10;

int t, n, m, num, k;

int mu[N], p[N], sum[N];

bool vis[N];

template<class T>inline void read(T &x) {

	x = 0; int f = 0; char ch = getchar();

	while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();

	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();

	x = f ? -x : x;

	return;

}

void get_mu(int n) {

	mu[1] = 1;

	for (int i = 2; i <= n; ++i) {

		if (!vis[i]) p[++num] = i, mu[i] = -1;

		for (int j = 1; j <= num; ++j) {

			if (i * p[j] > n) break;

			vis[i * p[j]] = 1;

			if (i % p[j] == 0) {

				mu[i * p[j]] = 0;

				break;

			} else mu[i * p[j]] = -mu[i];

		}

	}

}

int cal(int x, int y, int k) {

	int mx = min(x, y), ans = 0;

	for (int l = 1, r; l <= mx; l = r + 1) {

		r = min(x / (x / l), y / (y / l));

		ans += ((x / (l * k)) * (y / (l * k)) * (sum[r] - sum[l - 1]));

	}

	return ans;

}

int a, b, c, d;

signed main() {

	get_mu(N);

	for (int i = 1; i <= N; ++i) sum[i] = sum[i - 1] + mu[i];

	read(t);

	while (t --) {

		read(a), read(b), read(c), read(d), read(k);

		printf("%d\n", cal(b, d, k) - cal(b, c - 1, k) - cal(a - 1, d, k) + cal(a - 1, c - 1, k));

	}

	return 0;

}