人话：
两个长宽为整数的矩形，判断它们是否能包含。
思维姿势：一定是面积大的包含面积小的矩形（如果包含）。
假设面积小的矩形X长宽AB，面积大的矩形Y长宽CD。
大胆假设当X的三点都在Y上时，第四个点能决定了X能否“缩”在Y内。
枚举情况，并二分X“卡”在Y内的各种角度，判断第四个点是否在Y内。
真·罗干！
Code：
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
double _90=atan(1.0)*2;
double eps=1e-7;
double eps2=1e-3;
double f(double A,double B,double a)
{
return sin(a)*A+cos(a)*B;
}
double qia(double A,double B,double C,double l,double r)
{
if (f(A,B,l)<f(A,B,r))
{
while (r-l>=eps)
{
double mid=(l+r)/2.0;
if (f(A,B,mid)<C) l=mid;
else r=mid;
}
if (f(A,B,l)<=C+eps2)return l;
else return -1;
}
else
{
while (r-l>=eps)
{
double mid=(l+r)/2.0;
if (f(A,B,mid)<C) r=mid;
else l=mid;
}
if (f(A,B,l)<=C+eps2)return l;
else return -1;
}
}
int check(double A,double B,double C,double D)//A,B小
{
double m=atan(A/B);
double a1=qia(A,B,C,0,m);
if (a1!=-1) if (f(B,A,a1)<=D) return 1;
double a2=qia(A,B,C,m,_90);
if (a2!=-1) if (f(B,A,a2)<=D) return 1;
a1=qia(A,B,D,0,m);
if (a1!=-1) if (f(B,A,a1)<=C) return 1;
a2=qia(A,B,D,m,_90);
if (a2!=-1) if (f(B,A,a2)<=C) return 1;
return 0;
}
int main()
{
double A,B,C,D;
while (1)
{
scanf("%lf%lf%lf%lf",&A,&B,&C,&D);
if (check(A,B,C,D)||check(C,D,A,B)) puts("Yes");
else puts("No");
}
return 0;
}
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