【CUGBACM15级BC第四场 A】hdu 4931 Happy Three Friends

Happy Three Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1030    Accepted Submission(s): 764

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

31 2 3 3 2 22 2 2 2 2 21 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!
 

 

解题思路：

六个数排序，比较最大的两个数的和  与 第二三四三个数的和  比较

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;int a[105];int main(){    int t;    scanf("%d", &t);    while (t--)    {        memset(a, 0, sizeof(a));        for (int i = 0; i < 6; i++)        {            scanf("%d", &a[i]);        }        sort(a, a + 6, greater<int>());        int sum1 = a[0] + a[1];        int sum2 = a[2] + a[3] + a[4];        if (sum1 > sum2)        {            printf("Grandpa Shawn is the Winner!\n");        }        else        {            printf("What a sad story!\n");        }    }    return 0;}