bzoj2820-GCD

题意

\(T\le 10^4\) 次询问 \(n,m\) ，求

\[\sum _{i=1}^n\sum _{j=1}^m[gcd(i,j)\text { is prime}] \]

分析

这题还是很有趣的。设 \(n\le m\) 。

\[\begin{aligned} \sum _{i=1}^n\sum_{j=1}^m[gcd(i,j)\text { is prime}]&=\sum _{i=1}^n\sum _{j=1}^m\sum _k [k\text { is prime}][gcd(i,j)=k] \\ &=\sum _{i=1}^n\sum _{j=1}^m\sum _{k|i,k|j}[k\text { is prime}]\sum _{d|\frac{i}{k},d|\frac{j}{k}}\mu(d) \\ &=\sum _{d=1}^n\mu (d)\sum _{k=1}^n[k\text { is prime}]\sum _{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum _{j=1}^{\lfloor\frac{m}{k}\rfloor}[d|i,d|j] \\ &=\sum _{d=1}^n\mu (d)\sum _{k=1}^n[k\text { is prime}]\lfloor\frac{n}{kd}\rfloor \lfloor\frac{m}{kd}\rfloor \\ &=\sum _{i=1}^n\lfloor\frac{n}{i}\rfloor \lfloor\frac{m}{i}\rfloor\sum _{k|i,k\text { is prime}}\mu(\frac{i}{k}) \end{aligned} \]

令 \(f(x)=\sum _{k|x,k\text {is prime }}\mu (x/k)\) ，我们有：

\[ans=\sum _{i=1}^n\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{i}\rfloor f(i) \]

\(f(x)\) 可以在线性筛的过程中顺便处理出来，求前缀和就可以做到每次询问 \(O(\sqrt n)\) 。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long giant;
inline int read() {
	int x=0,f=1;
	char c=getchar_unlocked();
	for (;!isdigit(c);c=getchar_unlocked()) if (c=='-') f=-1;
	for (;isdigit(c);c=getchar_unlocked()) x=x*10+c-'0';
	return x*f;
}
const int maxn=1e7+1;
bool np[maxn];
int p[maxn],ps=0,mu[maxn],f[maxn];
int main() {
#ifndef ONLINE_JUDGE
	freopen("test.in","r",stdin);
#endif
	mu[1]=1,f[1]=0;
	for (int i=2;i<maxn;++i) {
		if (!np[i]) p[++ps]=i,mu[i]=-1,f[i]=1;
		for (int j=1,tmp;j<=ps && (tmp=i*p[j])<maxn;++j) {
			np[tmp]=true;
			if (i%p[j]) mu[tmp]=-mu[i],f[tmp]=mu[i]-f[i]; else {
				mu[tmp]=0;
				f[tmp]=mu[i];
				break;
			}
		}
	}
	for (int i=2;i<maxn;++i) f[i]+=f[i-1];
	int T=read();
	while (T--) {
		int n=read(),m=read();
		if (n>m) swap(n,m);
		giant ans=0;
		for (int i=1,j;i<=n;i=j+1) {
			j=min(n/(n/i),m/(m/i));
			ans+=(giant)(f[j]-f[i-1])*(n/i)*(m/i);
		}
		printf("%lld\n",ans);
	}
	return 0;
}