洛谷P2522 [HAOI2011]Problem b（莫比乌斯反演）

传送门

我们考虑容斥，设$ans(a,b)=sum_{i=1}^asum_{j=1}^b[gcd(a,b)==k]$，这个东西可以和这一题一样去算洛谷P3455 [POI2007]ZAP-Queries

然后只要在这上面加个容斥就好了，答案就是$ans(b,d)-ans(b,c-1)-ans(a-1,d)+ans(a-1,c-1)$

//minamoto
#include<iostream>
#include<cstdio>
#define ll long long
using namespace std;
#define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
inline int read(){
    #define num ch-'0'
    char ch;bool flag=0;int res;
    while(!isdigit(ch=getc()))
    (ch=='-')&&(flag=true);
    for(res=num;isdigit(ch=getc());res=res*10+num);
    (flag)&&(res=-res);
    #undef num
    return res;
}
const int N=5e5+5;
int p[N],m,vis[N],mu[N],sum[N],a,b,c,d,k;
void init(int n){
    mu[1]=1;
    for(int i=2;i<=n;++i){
        if(!vis[i]) p[++m]=i,mu[i]=-1;
        for(int j=1;j<=m&&p[j]*i<=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];
}
ll calc(int a,int b){
    int lim=min(a,b);ll res=0;
    for(int l=1,r;l<=lim;l=r+1){
        r=min(a/(a/l),b/(b/l));
        res+=1ll*(a/(l*k))*(b/(l*k))*(sum[r]-sum[l-1]);
    }
    return res;
}
int main(){
//    freopen("testdata.in","r",stdin);
    int T=read();init(50000);
    while(T--){
        a=read(),b=read(),c=read(),d=read(),k=read();
        printf("%lld
",calc(b,d)-calc(b,c-1)-calc(a-1,d)+calc(a-1,c-1));
    }
    return 0;
}