bzoj2301 莫比乌斯反演

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define N 60000
#define M 50000
int mindiv[N];
int zhi[N];
int zhin;
int u[N];
void shai(int x){
   for(int i=1;i<=x;i++)mindiv[i]=i;
   u[1]=1;
   for(int i=2;i<=x;i++){
       if(mindiv[i]==i)zhi[++zhin]=i,u[i]=-1;
       for(int j=1;j<=zhin&&zhi[j]<=mindiv[i]&&zhi[j]*i<=x;j++){
            mindiv[i*zhi[j]]=zhi[j];
            if(zhi[j]==mindiv[i])u[i*zhi[j]]=0;
            else u[i*zhi[j]]=-u[i];
        }    
    }    
}
int s[N];
int gcds(int a,int b,int k){
    if(a==0||b==0)return 0;
    int l,r=0;
    a/=k;
    b/=k;
    int ans=0;
    int lim=min(a,b);
    for(;;){
        l=r+1;
        if(l>lim)break;
        r=min((a/(a/l)),b/(b/l));
        ans+=(a/l)*(b/l)*(s[r]-s[l-1]);
    }    
    return ans;
}
int main(){
    shai(M);
    for(int i=1;i<=M;i++)s[i]=s[i-1]+u[i];
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
       int a,b,c,d,k;
       scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
       printf("%d\n",gcds(b,d,k)-gcds(b,c-1,k)-gcds(a-1,d,k)+gcds(a-1,c-1,k));    
    }
    
}

 

posted @ 2014-07-14 22:19  wangyucheng  阅读(...)  评论(...编辑  收藏