洛谷P2522 [HAOI2011]Problem b（莫比乌斯反演）

传送门

我们考虑容斥，设$ans(a,b)=\sum_{i=1}^a\sum_{j=1}^b[gcd(a,b)==k]$，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[<<],*p1=buf,*p2=buf;
 inline int read(){
     #define num ch-'0'
     char ch;bool flag=;int res;
     while(!isdigit(ch=getc()))
     (ch=='-')&&(flag=true);
     for(res=num;isdigit(ch=getc());res=res*+num);
     (flag)&&(res=-res);
     #undef num
     return res;
 }
 const int N=5e5+;
 int p[N],m,vis[N],mu[N],sum[N],a,b,c,d,k;
 void init(int n){
     mu[]=;
     for(int i=;i<=n;++i){
         if(!vis[i]) p[++m]=i,mu[i]=-;
         for(int j=;j<=m&&p[j]*i<=n;++j){
             vis[i*p[j]]=;
             if(i%p[j]==) break;
             mu[i*p[j]]=-mu[i];
         }
     }
     for(int i=;i<=n;++i) sum[i]=sum[i-]+mu[i];
 }
 ll calc(int a,int b){
     int lim=min(a,b);ll res=;
     for(int l=,r;l<=lim;l=r+){
         r=min(a/(a/l),b/(b/l));
         res+=1ll*(a/(l*k))*(b/(l*k))*(sum[r]-sum[l-]);
     }
     return res;
 }
 int main(){
 //    freopen("testdata.in","r",stdin);
     int T=read();init();
     while(T--){
         a=read(),b=read(),c=read(),d=read(),k=read();
         printf("%lld\n",calc(b,d)-calc(b,c-)-calc(a-,d)+calc(a-,c-));
     }
     return ;
 }