luogu2257 YY的GCD

题目链接

problem

给出一个\(n,m(n,m\le 10^7)\)，求\(\sum\limits_{i=1}^n\sum\limits_{j=1}^mgcd(i,j)\in P\)
P表示全部素数的集合。
\(T,(T\le 10000)\)组询问

solution

枚举因数

\[原式=\sum\limits_{d\in P} \sum\limits_{i=1}^{\frac{n}{d}}\sum\limits_{j=1}^{\frac{m}{d}}\epsilon(gcd(i,j)=1) \]

因为\(\epsilon=1*\mu\)

\[所以上式=\sum\limits_{d\in P} \sum\limits_{i=1}^{\frac{n}{d}}\sum\limits_{j=1}^{\frac{m}{d}}\sum\limits_{k|i,k|j}\mu(k)\\ =\sum\limits_{d\in P}\sum\limits_{k=1}^{\frac{min(n,m)}{d}}\mu(k)\lfloor\frac{n}{dk}\rfloor\lfloor\frac{m}{dk}\rfloor \]

令\(t=dk\)

原式=$$\sum\limits_{t=1}^{min(n,m)}\lfloor\frac{n}{t}\rfloor \lfloor\frac{m}{t}\rfloor\sum\limits_{d\in P,d|t}\mu(\frac{t}{d})$$

发现前面这一块数论分块就好了，后面就提前预处理出来，求个前缀和。

预处理复杂度\(O(nloglogn)\),单次询问复杂度\(O(\sqrt{n})\)

code

	/*
	* @Author: wxyww
	* @Date:   2020-01-20 19:08:26
	* @Last Modified time: 2020-01-20 19:18:39
	*/
	#include<cstdio>
	#include<iostream>
	#include<cstdlib>
	#include<cstring>
	#include<algorithm>
	#include<queue>
	#include<vector>
	#include<ctime>
	using namespace std;
	typedef long long ll;
	const int N = 10000010;
	ll read() {
		ll x=0,f=1;char c=getchar();
		while(c<'0'||c>'9') {
			if(c=='-') f=-1;
			c=getchar();
		}
		while(c>='0'&&c<='9') {
			x=x*10+c-'0';
			c=getchar();
		}
		return x*f;
	}
	int mu[N],pri[N],tot,vis[N];
	void pre() {
		mu[1] = 1;
		for(int i = 2;i < N;++i) {
			if(!vis[i]) pri[++tot] = i,mu[i] = -1;
			for(int j = 1;j <= tot && 1ll * i * pri[j] < N;++j) {
				vis[i * pri[j]] = 1;
				if(i % pri[j]) mu[i * pri[j]] = -mu[i];
				else {
					mu[i * pri[j]] = 0;break;
				}
			}
		}
	}
	int f[N];
	int main() {
		pre();

		for(int i = 1;i <= tot;++i) {
			int js = 1;
			for(int k = pri[i];k < N;k += pri[i],++js) {
				f[k] += mu[js];
			}
		}

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

		int T = read();
		while(T--) {
			int n = read(),m = read();
			ll ans = 0;
			for(int l = 1,r;l <= min(n,m);l = r + 1) {
				r = min(n / (n / l),m / (m / l));
				ans += 1ll * (n / l) * (m / l) * (f[r] - f[l - 1]);
			}
			printf("%lld\n",ans);
		}
		return 0;
	}