[莫比乌斯反演][整除分块] Bzoj P2301 Problem b

题目描述

对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数。

 

题解

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代码

#include <cstdio>
#include <iostream>
#define N 60010
#define ll long long
using namespace std;
ll n,k,mobius[N],p[N],bz[N],s[N];
void read(ll &x)
{
    x=0; ll p=1;
    static char c;c=getchar();
    while (!isdigit(c)){if(c=='-')p=-1;c=getchar();}
    while (isdigit(c)) {x=(x<<1)+(x<<3)+(c-48);c=getchar();}
    x*=p;   
}
ll calc(ll x,ll y)
{
    ll r=min(x,y),ans=0;
    for (ll a=1,b;a<=r;a=b+1) b=min(x/(x/a),y/(y/a)),ans+=((x/(a*k))*(y/(a*k))*(s[b]-s[a-1]));
    return ans;
}
int main()
{
    read(n),mobius[1]=1;
    for (ll i=2;i<=N;i++)
    {
        if (!bz[i]) mobius[i]=-1,p[++p[0]]=i;
        for (ll j=1;j<=p[0]&&i*p[j]<=N;j++)
        {
            bz[i*p[j]]=1;
            if (i%p[j]==0) break; else mobius[i*p[j]]=-mobius[i];
        }
    }
    for (ll i=1;i<=N;i++) s[i]=s[i-1]+mobius[i];
    for (ll a,b,c,d,k,l,r,sum;n;n--)
    {
        read(a),read(b),read(c),read(d),read(k);
        a=(a-1)/k,b=b/k,c=(c-1)/k,d=d/k,l=1,sum=0;
        if (a>c) swap(a,c),swap(b,d);
        for (ll A,B,C,D;l<=a;l=r+1) A=a/(a/l),B=b/(b/l),C=c/(c/l),D=d/(d/l),r=min(A,min(B,min(C,D))),sum+=(s[r]-s[l-1])*(b/l-a/l)*(d/l-c/l);
        for (ll A,B,C,D;l<=min(b,c);l=r+1) B=b/(b/l),C=c/(c/l),D=d/(d/l),r=min(B,min(C,D)),sum+=(s[r]-s[l-1])*(b/l)*(d/l-c/l);
        for (ll A,B,C,D;l<=min(b,d);l=r+1) B=b/(b/l),D=d/(d/l),r=min(B,D),sum+=(s[r]-s[l-1])*(b/l)*(d/l);
        printf("%lld\n",sum);
    }
}