Luogu P3579 [POI2014]PAN-Solar Panels

题目大意：

给出T组询问，每组询问给出四个数a,b,c,d，每次询问满足a<=x<=b,c<=y<=d的gcd(x,y)的最大值

首先可以想到如果存在gcd(x,y)=k，那么就一定要满足b/k>(a-1)/k&&d/k>(c-1)/k。

而n/k的k取法只有√n种，直接枚举即可。

//by zykykyk
#include<cstdio>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#define rg register
#define il inline
#define vd void
#define ll long long
#define For(i,x,y) for (rg int i=(x);i<=(y);i++)
#define Dow(i,x,y) for (rg int i=(x);i>=(y);i--)
#define cross(i,k) for (rg int i=first[k];i;i=last[i])
using namespace std;
il ll max(ll x,ll y){return x>y?x:y;}
il ll min(ll x,ll y){return x<y?x:y;}
il ll read(){
    ll x=0;int ch=getchar(),f=1;
    while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
    if (ch=='-'){f=-1;ch=getchar();}
    while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
int T,a,b,c,d,A,B,C,D,ans;
int main(){
    T=read();
    while (T--){
        a=read(),b=read(),c=read(),d=read(),ans=0;
        int l=max(max(sqrt(a-1),sqrt(b)),max(sqrt(c-1),sqrt(d)));
        For(i,1,l){
            int B=b/i,A=(a-1)/i,D=d/i,C=(c-1)/i;
            if (B>A&&D>C) ans=max(ans,i);
            if (A)
                if (b/A>(a-1)/A&&d/A>(c-1)/A) ans=max(ans,A);
            if (B)
                if (b/B>(a-1)/B&&d/B>(c-1)/B) ans=max(ans,B);
            if (C)
                if (b/C>(a-1)/C&&d/C>(c-1)/C) ans=max(ans,C);
            if (D)
                if (b/D>(a-1)/D&&d/D>(c-1)/D) ans=max(ans,D);
        }
        printf("%d
",ans);
    }
}