[4931]: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!
HintFor 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!

解题思路：这是一道简单的题目，其实大概意思就是：对于任意输入的6个整数，最大的两个最大数之和是否能够大于剩下4个数的任意三个数（take any of 3 numbers from the last（剩下） 4 numbers ）之和——–所以要与剩下的最大三个数之和比较

/*  author : Yangchengfeng*/#include<stdio.h>#define N 6int main(){    int t, num[N];    while(scanf("%d", &t)!=EOF){        while(t--){            int i = 0, j, max, temp, sumG = 0, sumB = 0;            for(; i<6; i++){                scanf("%d", &num[i]);            }            max = num[0];            for(i=0; i<6; i++){                for(j=i; j<6; j++){                    if(num[j] > num[i]){                        temp = num[j];                        num[j] = num[i];                        num[i] = temp;                    }                }            }            sumG = num[0] + num[1];            sumB = num[2] + num[3] + num[4];            if(sumG > sumB){                printf("Grandpa Shawn is the Winner!\n");            } else {                printf("What a sad story!\n");            }        }    }    return 0;}