C++ 解法, 执行用时: 2ms, 内存消耗: 476KB, 提交时间: 2022-04-21

#include<bits/stdc++.h>
using namespace std;
class rectangle{
	private:
		int length,width;
	public:
		void set(int x,int y){
			length=x;
			width=y;
		}
		int getlength(){
			return length;
		}
		int getwidth(){
			return width;
		}
		int area(){
			return length*width;
		}
		// write your code here......
    string cancover(rectangle &b)
    {
        if((length>=b.getlength()&&width>=b.getwidth())||(length>=b.getwidth()&&width>=b.getlength()))
            return "yes";
        else return "no";
    }
};
int main(){
	int l1,w1,l2,w2;
	cin>>l1>>w1>>l2>>w2;
	rectangle a,b;
	a.set(l1,w1);
	b.set(l2,w2);
	cout<<a.cancover(b);
	return 0;
}

C++ 解法, 执行用时: 3ms, 内存消耗: 384KB, 提交时间: 2022-05-17

#include<bits/stdc++.h>
using namespace std;
class rectangle{
	private:
		int length,width;
	public:
		void set(int x,int y){
			length=x;
			width=y;
		}
		int getlength(){
			return length;
		}
		int getwidth(){
			return width;
		}
		int area(){
			return length*width;
		}
		// write your code here......
        void cancover(rectangle a){
            if(this->area()>=a.area()){
                cout<<"yes";
            }
            else{
                cout<<"no";
            }
        }
};
int main(){
	int l1,w1,l2,w2;
	cin>>l1>>w1>>l2>>w2;
	rectangle a,b;
	a.set(l1,w1);
	b.set(l2,w2);
	a.cancover(b);
	return 0;
}

C++ 解法, 执行用时: 3ms, 内存消耗: 388KB, 提交时间: 2022-05-21

#include<bits/stdc++.h>
using namespace std;
class rectangle{
	private:
		int length,width;
	public:
		void set(int x,int y){
			length=x;
			width=y;
		}
		int getlength(){
			return length;
		}
		int getwidth(){
			return width;
		}
		int area(){
			return length*width;
		}
		// write your code here......
        string cancover(rectangle &r){
            if((this->getlength()>=r.getlength()&&this->getwidth()>=r.getwidth())||
              this->getlength()>=r.getwidth()&&this->getwidth()>=r.getlength())
                return "yes";
            else
                return "no";
            
            
            
        }
};
int main(){
	int l1,w1,l2,w2;
	cin>>l1>>w1>>l2>>w2;
	rectangle a,b;
	a.set(l1,w1);
	b.set(l2,w2);
	cout<<a.cancover(b);
	return 0;
}

C++ 解法, 执行用时: 3ms, 内存消耗: 392KB, 提交时间: 2022-06-08

#include<bits/stdc++.h>
using namespace std;
class rectangle {
  private:
    int length, width;
  public:
    void set(int x, int y) {
        length = x;
        width = y;
    }
    int getlength() {
        return length;
    }
    int getwidth() {
        return width;
    }
    int area() {
        return length * width;
    }
    // write your code here......
    string cancover(rectangle b) {
        int l1 = max(length, width);
        int w1 = min(length, width);
        int l2 = max(b.length, b.width);
        int w2 = min(b.length, b.width);
        if (l1 >= l2 && w1 >= w2) return "yes";
        else return "no";
    }
};
int main() {
    int l1, w1, l2, w2;
    cin >> l1 >> w1 >> l2 >> w2;
    rectangle a, b;
    a.set(l1, w1);
    b.set(l2, w2);
    
    cout << a.cancover(b);
    return 0;
}

C++ 解法, 执行用时: 3ms, 内存消耗: 392KB, 提交时间: 2022-04-02

#include<bits/stdc++.h>
using namespace std;
class rectangle{
	private:
		int length,width;
	public:
		void set(int x,int y){
			length=x;
			width=y;
		}
		int getlength(){
			return length;
		}
		int getwidth(){
			return width;
		}
		int area(){
			return length*width;
		}
		// write your code here......
    string cancover(rectangle b){
        if(length>=(b.getlength()||width>=b.getwidth())&&length*width>=b.area())
            return "yes";
        else
            return "no";
    }
    
};
int main(){
	int l1,w1,l2,w2;
	cin>>l1>>w1>>l2>>w2;
	rectangle a,b;
	a.set(l1,w1);
	b.set(l2,w2);
	cout<<a.cancover(b);
	return 0;
}