ZAP-Queries【POI2007】【莫比烏斯反演】
阿新 • • 發佈:2018-12-15
傳送門:https://www.luogu.org/problemnew/show/P3455
太經典了,,模板往上套
我直接上個截圖吧,,打公式太麻煩
整除分塊也是常識,直接上啦
#include<bits/stdc++.h> #define in read() using namespace std; inline int in{ int cnt=0,f=1; char ch=0; while(!isdigit(ch)){ ch=getchar(); if(ch=='-')f=-1; } while(isdigit(ch)){ cnt=cnt*10+ch-48; ch=getchar(); } return cnt*f; } #define N 50003 int mu[N],prim[N],cnt,vis[N]; long long sum[N]; inline void mumu(int n){ mu[1]=1; for(register int i=2;i<=n;i++){ if(!vis[i]){ prim[++cnt]=i; mu[i]=-1; } for(register int j=1;j<=cnt&&prim[j]*i<=n;j++){ vis[prim[j]*i]=1; if(i%prim[j]==0)break; else mu[prim[j]*i]=-mu[i]; } } for(register int i=1;i<=n;i++)sum[i]=sum[i-1]+(long long)mu[i]; }int a,b,c,d,k; inline long long query(int a,int b){ if(a>b)swap(a,b);long long ans=0; for(register int l=1,r;l<=a;l=r+1){ r=min(a/(a/l),b/(b/l)); ans+=(1ll*a/(1ll*l*k))*(1ll*b/(1ll*l*k))*(sum[r]-sum[l-1]); } return ans; } int main(){ int t=in;mumu(N); while(t--){ a=in;b=in;k=in; printf("%lld\n",query(a,b)); } return 0; }