鸡兔同笼

题目很简单，陷阱还挺多的，首先鸡or兔数目要>=0 ,还有腿的数目必须为偶数个

#include<iostream>
using namespace std;
int main()
{
int N;
cin>>N;
while(N--)
{
int NUM,leg,chicken,rabit;
cin>>NUM>>leg;
chicken=(4*NUM-leg)/2;
rabit=NUM-chicken;
if((chicken>=0&&chicken<=NUM)&&leg%2==0)
cout<<chicken<<" "<<rabit<<endl;
else
cout<<"No answer"<<endl;

}

}