数字游戏。

思路：

根据n，max，min来确定平均值的范围即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long


int main()
{
    
    
	int t;
	cin>>t;
	while(t--)
	{
    
    
		int n;
		cin>>n;
		double maxx,minn,ave;
		cin>>maxx>>minn>>ave;
		if(maxx < minn || ave < minn || ave > maxx)
		{
    
    
			cout<<"no"<<endl;
			continue;
		}
		maxx = (n-1)*maxx+minn,maxx/=n;
		minn = (n-1)*minn+maxx,minn/n;
		if(ave >= minn && ave <= maxx)cout<<"yes"<<endl;
	}
}