链接：https://vjudge.net/contest/333183#problem/D

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. 

Input

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.

Output

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

Sample Input

3
1 1 2
2 4 2
3 3 3

Sample Output

Alice 0 Bob 1
Alice 1 Bob 1
Alice 2 Bob 1

思路：

作为一名蒟蒻，看着题解也写出tle，QAQ，后来还好有老瞎眼学长独具慧眼看出了我tle的原因。首先这道题是一道看似博弈的dp,也就是博弈dp，我们无法得知每次拍卖时对方拍卖的价格，所以对于A或B来说，他可能使用任意的钱来拍卖当前拍品，但我们可以得知对方还剩余多少钱（似乎没什么用QAQ）

题解：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
typedef long long ll;
using namespace std;
int dp[256][256][256];
int f(int n,int a)
{
    int t=n/2;
    return t*a+(n-t)*(a+1);
}
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++)
	     {
	        int t=min(f(i,a),256);
	        for(int b=0;b<t;b++)
	        {
	            if(a==b)
				{
					dp[i][a][b]=(i+1)/2; 
					continue;
				}  
	            int ta=i-dp[i-1][b][a];
	            int tb=0;
	            int j=1;
	            while(1)
	            {
					tb=i-1-dp[i-1][b-j][a];
	                if(b<j||tb>=ta) 
	                {
	                    dp[i][a][b]=ta;
	                    break;
	                }	
					ta=i-dp[i-1][b][a-j];
	                if(a<j||ta<=tb)  
	                {
	                    dp[i][a][b]=tb;
	                    break;
	                }
	                j++; 
	            }
	        }
    	}
    }       
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,a,b;
        scanf("%d%d%d",&n,&a,&b);
        printf("Alice %d Bob %d\n",dp[n][a][b],n-dp[n][a][b]);
    }
    return 0;
}