BZOJ 2301 [HAOI2011]Problem b (莫比乌斯反演)

题目大意：已知xin [a,b],yin [c,d]，求gcd(x,y)为k的有序数对数量 a,b,c,d,k及询问数<=50000

比yy的gcd好做一些吧

转化题目，直接求解比较困难，利用容斥原理，问题转化为求$ans(b,d)-ans(a-1,d)-ans(b,c-1)+ans(a-1,c-1)$

求$sum_{i=1}^{n}sum_{j=1}^{m}[gcd(i,j)==k]$

化简$sum_{i=1}^{left lfloor frac{n}{k} ight floor}sum_{j=1}^{left lfloor frac{m}{k} ight floor}[gcd(i,j)==1]$

套路变形$sum_{d=1}^{left lfloor frac{n}{k} ight floor}left lfloor frac{n}{kd} ight floor cdot left lfloor frac{m}{kd} ight floorcdot mu(d)$

预处理$mu(d)$的前缀和，再整除分块的思想求解即可

与yy的gcd那道题不同，整除分块的部分有一些细节需要思考

#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 50100
#define maxn 50000
#define ll long long 
using namespace std;

int T,a,b,c,d,K,cnt;
int use[N],pr[N],mu[N],pmu[N];
void Pre()
{
    mu[1]=1;
    for(int i=2;i<=maxn;i++)
    {
        if(!use[i]) pr[++cnt]=i,mu[i]=-1;
        for(int j=1;j<=cnt&&i*pr[j]<=maxn;j++){
            use[i*pr[j]]=1;
            if(i%pr[j]==0){    
                mu[i*pr[j]]=0;
                break;
            }else{
                mu[i*pr[j]]=-mu[i];
            }
        }
    }
    for(int i=1;i<=maxn;i++)
        pmu[i]=pmu[i-1]+mu[i];
}
ll solve(int n,int m)
{
    if(n>m) swap(n,m);
    ll ans=0;int nd,md;
    for(int i=1,la,mi=min(n/K,m/K);i<=mi;i=la+1)
    {
        nd=n/K,md=m/K;
        la=min(nd/(nd/i),md/(md/i));
        ans+=1ll*(n/(K*i))*(m/(K*i))*(pmu[la]-pmu[i-1]);
    }return ans;
}
int main()
{
    scanf("%d",&T);
    Pre();
    for(int t=1;t<=T;t++)
    {
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&K);
        printf("%lld
",solve(b,d)-solve(a-1,d)-solve(b,c-1)+solve(a-1,c-1));
    }
    return 0;
}