Forever Young

洛谷 P3455 [POI2007]ZAP-Queries

题意

\(T\)组询问,每次询问求

\[\sum\limits_{x=1}^{a}\sum\limits_{y=1}^{b}[\gcd(x,y)=d] \]

思路

因为我不喜欢用\(x、y、a、b、d\),所以一一对应换成\(i、j、n、m、k\)

直接淦式子

\[\begin{align*}&\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[\gcd(i,j)=k]\\=& \sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}[\gcd(i,j)=1]\\=&\sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}\sum\limits_{d|\gcd(i,j)}\mu(d)\\=&\sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}\sum\limits_{d|i}\sum _{d|j}\mu(d)\\=&\sum\limits_{d=1}^{\min(\lfloor\frac nk\rfloor,\lfloor\frac mk\rfloor)}\mu(d)\sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}\sum\limits_{d|i}\sum\limits_{d|j}1\\=&\sum\limits_{d=1}^{\min(\lfloor\frac nk\rfloor,\lfloor\frac mk\rfloor)}\mu(d)\sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{d|i}1\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}\sum\limits_{d|j}1\\=&\sum\limits_{d=1}^{\min(\lfloor\frac nk\rfloor,\lfloor\frac mk\rfloor)}\mu(d)\lfloor\frac{\lfloor \frac nk\rfloor}d\rfloor\lfloor\lfloor\frac{\frac mk\rfloor}d\rfloor\\=&\sum\limits_{d=1}^{\min(\lfloor\frac nk\rfloor,\lfloor\frac mk\rfloor)}\mu(d)\lfloor\frac n{kd}\rfloor\lfloor\frac m{kd}\rfloor\end{align*} \]

现在就可以每次询问\(O(n)\)做这道题了

但是跑不过啊,不过显然可以数论分块,所以我们就可以\(O(\sqrt n)\)回答每次询问了

代码

/*
Author:loceaner
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int n, m, k, mu[A], p[A], sum[A], cnt;
bool vis[A];

void getmu(int n) {
	mu[1] = 1;
	for (int i = 2; i <= n; i++) {
		if (!vis[i]) p[++cnt] = i, mu[i] = -1;
		for (int j = 1; j <= cnt && i * p[j] <= n; j++) {
			vis[i * p[j]] = 1;
			if (i % p[j] == 0) break;
			mu[i * p[j]] -= mu[i];
		}
	}
	for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i];
}

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

int main() {
	getmu(50000);
	int T = read();
	while (T--) {
		n = read(), m = read(), k = read();
		cout << solve(n, m, k) << '\n';
	}
	return 0;
}
posted @ 2020-04-28 09:59  Loceaner  阅读(136)  评论(4编辑  收藏  举报