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BestCoder Round #4(Happy Three Friends-贪心)

2024-07-17 19:05:00   218人阅读


Happy Three Friends


Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 934 Accepted Submission(s): 738


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!

直接把最大2个放外面，Beautiful-leg Mzry剩下的中再取最大的3个


#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN ()
#define MAXM ()
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int a[7],t;
int main()
{
//	freopen("a.in","r",stdin);
//	freopen("a.out","w",stdout);
	
	cin>>t;
	while (t--)
	{
		For(i,6) cin>>a[i];
		sort(a+1,a+7);
		if (a[6]+a[5]>a[4]+a[3]+a[2]) cout<<"Grandpa Shawn is the Winner!"<<endl;
		else cout<<"What a sad story!"<<endl;		
	}		
	return 0;
}





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