Happy Three Friends
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 70 Accepted Submission(s): 62
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
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。
代码：
#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;
}
[BestCoder Round #4] hdu 4931 Happy Three Friends