Happy Three Friends

Problem Description

Dong-hao , Grandpa Shawn , Beautful-leg Mzry are good friends. One day , they want to play a game.

There are 6 numbers on the table. 

Firstly , Dong-hao can change the order of 6 numbers.

Secondly , Grandpa Shawn take the first one and the last one , sum them up as his scores.

Thirdly , Beautiful-leg Mzry take any of 3 numbers from the last 4 numbers , and sum them up as his scores.

Finally , if Grandpa Shawn‘s score is larger than Beautiful-leg Mzry‘s , Granpa Shawn wins! 

If Grandpa Shawn‘s score is smaller than Beautiful-leg Mzry‘s , Granpa Shawn loses.

If the scores are equal , there is a tie.

Nowadays , it‘s really sad that Grandpa Shawn loses his love. So Dong-hao wants him to win(not even tie). You have to tell Dong-hao whether he can achieve his goal.

 

Input

There is a number T shows there are T test cases below. ( T <= 50)

For each test case , there are 6 numbers Ai ( 1 <= Ai <= 100 ).

 

Output

If Dong-hao can achieve his goal , output "Grandpa Shawn is the Winner!"
If he can not , output "What a sad story!"

 

Sample Input

3
1 2 3 3 2 2
2 2 2 2 2 2
1 2 2 2 3 4

 

Sample Output

What a sad story!
What a sad story!
Grandpa Shawn is the Winner!
Hint
For the first test case , {3 , 1 , 2 , 2 , 2 , 3} Grandpa Shawn can take 6 at most . But Beautiful-leg Mzry can take 6 too. So there is a tie.
For the second test cases , Grandpa Shawn loses.
For the last one , Dong-hao can arrange the numbers as {3 , 2 , 2 , 2 , 1 , 4} , Grandpa Shawn can take 7 , but Beautiful-leg Mzry can take 6 at most. So Grandpa Shawn Wins!
 

 

Source

BestCoder Round #4

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <iomanip>
using namespace std;
int num[8];

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        for(int i=1;i<=6;i++)
            scanf("%d",&num[i]);
        sort(num+1,num+7);
        int dong=num[5]+num[6];
        int other=num[4]+num[3]+num[2];
        if(dong>other)
            cout<<"Grandpa Shawn is the Winner!"<<endl;
        else
            cout<<"What a sad story!"<<endl;
    }
    return 0;
}