Alice and Bob play an auction game. Alice has A dollars and Bob has B dollars initially. There are N items on sale. In each round, an item will be sold by the following way. Alice writes down an integer a (0 ≤ a ≤ A) and Bob writes down an integer b (0 ≤ b ≤ B), which are the amount of dollars they want to pay for the item. If a > b, then Alice gets the item and pays a dollars to the seller. If a < b, then Bob gets the item and pays b dollars to the seller. If a = b, then for the 1st, 3rd, 5th, 7th … round, Alice gets the item and pays a dollars; for the 2rd, 4th, 6th, 8th … round, Bob gets the item and pays b dollars. Since all the items have the same value, the goal of the auction game is to get as many items as possible. Both Alice and Bob know the values of N,A and B. Your task is to calculate how many items they will get if both of them play optimally.

输入
The first line is the number of test cases. Each test case contains 3 integers N,A and B, which are no larger than 255.

输出
For each test case, output the number of items Alice and Bob will get if both of them play optimally.

样例输入 Copy
3
1 1 2
2 4 2
3 3 3

样例输出 Copy
Alice 0 Bob 1
Alice 1 Bob 1
Alice 2 Bob 1

【题意】
t组样例，每组给n轮，一人有a钱，一人有b钱，问最后结果

【思路】
博弈的重点是，每次对新的一句博弈a，b分别拿出多少来争取新的这局，最后结果如何
显然拿出东西来博弈，结果还不如不拿，那就没必要继续拿东西出来了

#pragma GCC optimize(3,"Ofast","inline")
#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
int dp[256][256][256],t,n,x,y;
int main()
{
    
    for(int i=0;i<256;i++)
        for(int j=0;j<256;j++){
    
            if(i>=j)dp[1][i][j]=1;
            else  dp[1][i][j]=0;
        }
    for(int i=2;i<256;i++)
        for(int a=0;a<256;a++)
            for(int b=0;b<256;b++){
    
                if(i&1){
    
                    int va=dp[i-1][a][b]+1,vb;
                    for(int add=1;;add++){
    
                        if(add>b||(vb=dp[i-1][a][b-add])>=va){
    
                            dp[i][a][b]=va;
                            break;
                        }
                        if(add>a||(va=dp[i-1][a-add][b]+1)<=vb){
    
                            dp[i][a][b]=vb;
                            break;
                        }
                    }
                }else
                {
    
                    int vb=dp[i-1][a][b],va;
                    for(int add=1;;add++){
    
                        if(add>a||(va=dp[i-1][a-add][b]+1)<=vb){
    
                            dp[i][a][b]=vb;
                            break;
                        }
                        if(add>b||(vb=dp[i-1][a][b-add])>=va){
    
                            dp[i][a][b]=va;
                            break;
                        }
                    }
                }
            }
    scanf("%d",&t);
    while(t--)
    {
    
        scanf("%d%d%d",&n,&x,&y);
        printf("Alice %d Bob %d\n",dp[n][x][y],n-dp[n][x][y]);
    }
    return 0;
}