[HNOI2011]Problem B

Description:

给定\(a\)，\(b\)，\(c\)，\(d\)，\(k\)
求:
\(\sum_{i=a}^{b} \sum_{j=c}^{d} gcd(i,j)==k\)
\(T\)组询问

Hint:

均\(<=5e4\)

Solution:

不就比\(zap-query\)多了个下界嘛，容斥即可

#include<bits/stdc++.h>
using namespace std;
const int mxn=1e5+5;
int T,tot,mu[mxn],vis[mxn],p[mxn],f[mxn];
int sum[mxn];

void sieve(int lim)
{
    mu[1]=1;
    for(int i=2;i<=lim;++i) {
        if(!vis[i]) mu[i]=-1,p[++tot]=i;
        for(int j=1;j<=tot;++j) {
            if(p[j]*i>lim) break;
            vis[p[j]*i]=1;
            if(i%p[j]==0) {
                mu[i*p[j]]=0;
                break;
            }
            mu[p[j]*i]=-mu[i];
        }
    }
    for(int i=1;i<=lim;++i) sum[i]=sum[i-1]+mu[i];
}

int solve(int n,int m,int k)
{
    if(n>m) swap(n,m); int ans=0;
    for(int l=1,r;l<=n;l=r+1) {
        r=min(n/(n/l),m/(m/l));
        ans+=1ll*(sum[r]-sum[l-1])*(n/l)*(m/l);
    }
    return ans;
}

int main()
{
    scanf("%d",&T);
    sieve(50000);
    while(T--) {
        int n,m,x,y,k;
        scanf("%d%d%d%d%d",&n,&m,&x,&y,&k);  --n,--x;
        n/=k,m/=k,x/=k,y/=k;
        printf("%d\n",solve(m,y,k)-solve(n,y,k)-solve(m,x,k)+solve(n,x,k));
    }
    return 0;
}