【BZOJ】2301 Problem b

Problem b

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题目链接

• Problem b

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题目大意

    给你两个区间[a,b]和[c,d]，让你求有多少个(x,y)满足:

gcd(x,y)=k且a<=x<=b、c<=y<=d

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题解

莫比乌斯反演

    首先用容斥原理把问题分解成求：

[1,n]、[1,m]中有多少个gcd(x,y)=1

    然后就是我们上一篇解决的问题了，具体可以看这里，上一篇写的很详细：莫比乌斯反演——BZOJ 2818

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代码

#include <iostream>#include <cstring>#include <cstdio>#define LL long long#define maxn 50005using namespace std;int a,b,c,d,n,k,p[maxn],mu[maxn];int cnt;bool vis[maxn];void setup(int high){    mu[1]=1; cnt=0;    for (int i=2;i<=high;i++)    {        if (!vis[i])        {            vis[i]=1; mu[i]=-1;            p[cnt++]=i;        }        for (int j=0;j<cnt && i*p[j]<=high;j++)        {            vis[i*p[j]]=1;            if (i%p[j]) mu[i*p[j]]=-mu[i];            else            {                mu[i*p[j]]=0;                break;            }        }    }    for (int i=1;i<=high;i++) mu[i]+=mu[i-1];}int solve(int n,int m){    int p,ans=0,last=0;    n/=k; m/=k;    p=min(n,m);    for (int i=1;i<=p;i=last+1)    {        last=min(n/(n/i),m/(m/i));        ans+=(mu[last]-mu[i-1])*(n/i)*(m/i);    }    return ans;}int main(){    setup(maxn-5);    int T;    scanf("%d",&T);    while (T--)    {        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);        printf("%d\n",solve(b,d)-solve(a-1,d)-solve(b,c-1)+solve(a-1,c-1));    }    return 0;}