BestCoder Round #4(Happy Three Friends-贪心)

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

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!

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;}