#include <iostream>
using namespace std;
int main(){
    int n,m,num;
    cin>>n>>m;
    num=n;
    n*=4;
    if(m>n||(n-m)%2==1||((n-m)/2)>num){
        cout<<"No answer"<<endl;
        return 0;
    }
    m=(n-m)/2;
    cout<<m<<" "<<num-m<<endl;
    return 0;
}

其他人在干什么？没一个代码是能AC又简洁易懂的！ 干货来了！ 已验证，保证AC，不AC我再也不学c++了，放心复制↓

#include <iostream>
using namespace std;
int main()
{
	int n,m,c,r;
	cin >> n >> m;
	r = (m - 2 * n) / 2;
	c = n - r;
	if (r < 0 || c < 0 || (m - 2 * n) % 2 != 0)
		cout << "No answer";
	else
		cout << c << endl << r;
	 
    return 0;
}

极简代码

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,INF=0x3f3f3f3f;
int n,m; 
int main()
{
	cin>>n>>m;
	if(m<n*2||(m-n*2)%2!=0||(m-n*2)/2>n)cout<<"No answer";
	else printf("%d %d",n-((m-n*2)/2),(m-n*2)/2);
	return 0;
}

只有两行（doge #include<bits/stdc++.h> using namespace std;int n,m,ans;int main(){cin>>n>>m;if((m%2==0)&&((n*4-m)/2>=0)&&((n*4-m)/2<=n))ans++;if(ans==1){cout<<(n*4-m)/2<<endl<<n-(n*4-m)/2;}else{cout<<"No answer";return 0;} 这题特殊情况有点多

#include <iostream>
using namespace std;
int main(){
    int n,m,num;
    cin>>n>>m;
    num=n;
    n*=4;
    if(m>n||(n-m)%2==1||((n-m)/2)>num){
        cout<<"No answer"<<endl;
        return 0;
    }
    m=(n-m)/2;
    cout<<m<<" "<<num-m<<endl;
    return 0;
}

n,m=input().split()

a=round((int(m)-int(n)*2)/(4-2))

if int(n)-a <0 or a<0:

print('No answer') 

elif int(m)<int(n) or int(m)%2==1:

print('No answer') 

else:

print(int(n)- a,a)

n,m=input().split() a=round((int(m)-int(n)*2)/(4-2)) if int(n)-a <0 or a<0: print('No answer') elif int(m)<int(n) or int(m)%2==1: print('No answer') else: print(int(n)- a,a)