【Description】

In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl 
named "Sena" are playing a video game. The game system of this video game is quite 
unique: in the process of playing this game, you need to constantly face the 
choice, each time you choose the game will provide 1
 
−31-31−3 options, the player 
can only choose one of them. Each option has an effect on a "score" parameter in 
the game. Some options will increase the score, some options will reduce the score, 
and some options will change the score to a value multiplied by −1-1−1 .
That is, if there are three options in
 
 a selection, the score will be increased by 
111, decreased by 111, or multiplied by −1-1−1. The score before the selection is 
888. Then selecting option 111 will make the score become 999, and selecting option
222 will make the score 777 and select option 333 to make the score −8-8−8. Note 
that the
 
 score has an upper limit of 100100100 and a lower limit of −100-100−100. 
If the score is 999999 at this time, an option that makes the score +2+2+2 is 
selected. After that, the score will change to 100100100 and vice versa .
After all the choices have been made, the score will affect the ending of the game. 
If the score is greater than or equal to a certain value kkk, it will enter a good 
ending; if it is less than or equal to a certain value lll, it will enter the bad 
ending; if both conditions are not satisfied, it will enter the normal ending. Now,
Koutarou and Sena want to play the good endings and the bad endings respectively. 
They refused to give up each other and finally decided to use the "one person to 
make a choice" way to play the game, Koutarou first choose. Now assume that they 
all know the initial score, the impact of each option, and the kkk, lll values, and 
decide to choose in the way that works best for them. (That is, they will try their 
best to play the ending they want. If it's impossible, they would rather normal 
ending than the ending their rival wants.)
Koutarou and Sena are playing very happy, but I believe you have seen through the 
final ending. Now give you the initial score, the kkk value, the lll value, and the 
effect of each option on the score. Can you answer the final ending of the game?

【Input】

The first line contains four integers n,m,k,ln,m,k,ln,m,k,l（1≤n≤10001\le n \le 
10001≤n≤1000, −100≤m≤100-100 \le m \le 100−100≤m≤100 , −100≤l<k≤100-100 \le l < k 
\le 100−100≤l<k≤100 ）, represents the number of choices, the initial score, the 
minimum score required to enter a good ending, and the highest score required to 
enter a bad ending, respectively.
Each of the next nnn lines contains three integers a,b,ca,b,ca,b,c（a≥0a\ge 0a≥0 ,
b≥0b\ge0b≥0 ,c=0c=0c=0 or c=1c=1c=1）,indicates the options that appear in this 
selection,in which a=0a=0a=0 means there is no option to increase the score in this 
selection, a>0a>0a>0 means there is an option in this selection to increase the 
score by aaa ; b=0b=0b=0 means there is no option to decrease the score in this 
selection, b>0b>0b>0 means there is an option in this selection to decrease the 
score by bbb; c=0c=0c=0 means there is no option to multiply the score by −1-1−1 in 
this selection , c=1c=1c=1 means there is exactly an option in this selection to 
multiply the score by −1-1−1. It is guaranteed that a,b,ca,b,ca,b,c are not equal 
to 000 at the same time.

【Output】

 One line contains the final ending of the game. If it will enter a good 
ending,print "Good Ending"(without quotes); if it will enter a bad ending,print 
"Bad Ending"(without quotes);otherwise print "Normal Ending"(without quotes).

【Examples】

Sample Input

3 -8 5 -5
3 1 1
2 0 1
0 2 1

Sample Output

Good Ending

Sample Input

3 0 10 3
0 0 1
0 10 1
0 2 1

Sample Output

Bad Ending

【Problem Description】

兩個人玩遊戲，有n個輪次，初始分數為m，每個輪次有三種選擇，將m增加a，將m減少b，如果c==1,可以將m取
相反數。只能選擇一個對m進行操作。K玩家想要使最終分數大於等於k，S玩家想要使最終分數小於等於l，K玩家
先手，問最終K玩家能否使分數大於等於k。

【Solution】

博弈論+dp，記憶化搜尋。（挺好的題）
定義dp[i][j]為第i個輪次，分數為j時，所能得到的最優分數為dp[i][j]
n個輪次，0~n-1，並且K玩家先手，所以進行搜尋時，輪次t為偶數時，求最大值，否則求最小值。
因為分數可能為負數，所以增加100的偏移量，便於儲存。

【Code】

/*
 * @Author: Simon 
 * @Date: 2018-09-09 19:35:20 
 * @Last Modified by: Simon
 * @Last Modified time: 2018-09-09 20:41:00
 */
#include<bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 1005

int n, m, k, l;
struct node
{
    int a,b,c;
    node(){}
    node(int _a,int _b,int _c):a(_a),b(_b),c(_c){}
}p[maxn];
int dp[maxn][205];//dp[i][j]為第i個輪次，分數為j時，所能得到的最優分數為dp[i][j]

int dfs(int t,int score)
{
    if(t>=n) return score-100;
    if(dp[t][score]!=INF) return dp[t][score];
    if(t%2==0)//K玩家
    {
        int ans=score-100,tmp=-100;
        if(p[t].a>0)
        {
            int n_ans=min((int)100,ans+p[t].a);
            tmp=max(tmp,dfs(t+1,n_ans+100));
        }
        if(p[t].b>0)
        {
            int n_ans=max((int)-100,ans-p[t].b);
            tmp=max(tmp,dfs(t+1,n_ans+100));
        }
        if(p[t].c>0)
        {
           tmp=max(tmp,dfs(t+1,-ans+100));
        }
        dp[t][score]=tmp;
        return tmp;
    }
    else//S玩家
    {
        int ans=score-100,tmp=100;
        if(p[t].a>0)
        {
            int n_ans=min((int)100,ans+p[t].a);
            tmp=min(tmp,dfs(t+1,n_ans+100));
        }
        if(p[t].b>0)
        {
            int n_ans=max((int)-100,ans-p[t].b);
            tmp=min(tmp,dfs(t+1,n_ans+100));
        }
        if(p[t].c>0)
        {
            tmp=min(tmp,dfs(t+1,-ans+100));
        }
        dp[t][score]=tmp;
        return tmp;
    }
}
Int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>m>>k>>l;
    for(int i=0;i<maxn;i++) for(int j=0;j<205;j++) dp[i][j]=INF;
    for(int i=0;i<n;i++) cin>>p[i].a>>p[i].b>>p[i].c;

    int ans=dfs(0,m+100);
    if(ans>=k) puts("Good Ending");
    else if(ans<=l) puts("Bad Ending");
    else puts("Normal Ending");

    cin.get(),cin.get();
    return 0;
}