[POI2007]ZAP-Queries

洛咕

题意:给定(a,b,d),求(sum_{i=1}^{a}sum_{j=1}^{b}(gcd(i,j)==d))

分析:莫比乌斯反演在(gcd)问题上的运用都有一定的套路.设,

(f(d)=sum_{i=1}^{a}sum_{j=1}^{b}(gcd(i,j)==d))

(F(d)=sum_{d|x}^{n}f(x))

仔细研究一下(F(d))的式子,不难发现(F(d))就是指在1到(a)和1到(b)中各选1个数,它们的(gcd)是(d)的倍数的方案数，那么显然有式子(F(d)=lfloorfrac ad floorlfloorfrac bd floor)

又由莫比乌斯反演定理可得,(f(d)=sum_{d|x}μ(frac xd)F(x))

代入(F(x)=lfloorfrac ax floorlfloorfrac bx floor)得,(f(d)=sum_{d|x}μ(frac xd)lfloorfrac ax floorlfloorfrac bx floor)

故(ans=f(d)=sum_{d|x}μ(frac xd)lfloorfrac ax floorlfloorfrac bx floor=sum_{x=1}^{min(a/d,b/d)}μ(frac xd)lfloorfrac ax floorlfloorfrac bx floor)

预处理出(μ)函数的前缀和,然后整除分块就好了.

#include<bits/stdc++.h>
using namespace std;
inline int read(){
   int s=0,w=1;char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}
   return s*w;
}
const int N=50005;
int mu[N],prime[N],v[N],sum[N];
inline void get_mu(){
    int m=0;mu[1]=1;
    for(int i=2;i<=N;i++){
		if(!v[i]){
	    	prime[++m]=i;
	    	v[i]=i;
	    	mu[i]=-1;
		}
		for(int j=1;j<=m;j++){
	    	if(i*prime[j]>N)break;
	    	v[i*prime[j]]=prime[j];
	    	if(i%prime[j]==0)break;
	    	else mu[i*prime[j]]=-mu[i];
		}
    }
    for(int i=1;i<=N;i++)sum[i]=sum[i-1]+mu[i];
}
int main(){
    get_mu();
    int n=read();
    while(n--){
		int a=read(),b=read(),d=read();
		a/=d;b/=d;int ans=0;
		for(int l=1,r;l<=min(a,b);l=r+1){
	    	r=min(a/(a/l),b/(b/l));
	    	ans+=(a/l)*(b/l)*(sum[r]-sum[l-1]);
		}
		printf("%d
",ans);
    }
    return 0;
}