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problem=1199" style="color:rgb(79,107,114)">Statistics

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Alice and Bob are playing a strange game. The rules of the game are:

1. Initially there are n piles.

2. A pile is formed by some cells.

3. Alice starts the game and they alternate turns.

4. In each tern a player can pick any pile and divide it into two unequal piles.

5. If a player cannot do so, he/she loses the game.

Now you are given the number of cells in each of the piles, you have to find the winner of the game if both of them play optimally.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 100). The next line contains n integers, where the i^th integer denotes the number of cells in the i^th pile. You can assume that the number of cells in each pile is between 1

Output

For each case, print the case number and ‘Alice‘ or ‘Bob‘ depending on the winner of the game.

Explanation

In case 1, Alice has only 1 move, she divides the pile with 4 cells into two unequal piles, where one pile has 1 cell and the other pile has 3 cells. Now it‘s Bob‘s turn. Bob divides the pile with 3 cells into two piles, where one pile has 1 cell and another pile has 2 cells. So, now there are three piles having cells 1, 1, 2. And Alice loses, since she doesn‘t have any moves now.

有n堆石子(1<=n<=100)，每一堆分別有ai個石子(1<=ai<=10000)，一次操作能夠使一堆石子變成兩堆數目不相等（註意是不相等）的石子，最後不能操作就算輸，問先手贏還是後手贏。

解題思路:

就是一個SG函數，提到SG函數這個就是求一下 當前狀態的下一個狀態，又由於 這 n 堆石子是相互獨立的，沒有影響 所以說 能夠開用SG函數，

依據SG定理，如果 當前堆中有 m塊石子 那麽他的下一狀態就可能有 {1,m-1},{2,n-2},...,{(m-1)/2,m-(m-1)/2}（把每一種情況都想到 而且分析出來）。

然後分完的那些 a和b塊石子又能夠進行分，以此類推。那麽SG(x) = mex{ SG(1)^SG(x-1), SG(2)^SG(x-2),..., SG((x-1)/2)^SG(x-(x-1)/2) }，

然後我們要求的就是 SG[a[0]]^SG[a[1]]^...^SG[a[n-1]],假設結果是0就是 後手贏，否則 先手贏

My Code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 10000+5;
int sg[MAXN];
int hash[MAXN];
void Get_sg()///模板
{
    memset(sg, 0, sizeof(sg));
    for(int i=1; i<MAXN; i++)
    {
        memset(hash, 0, sizeof(hash));
        for(int j=1; j*2<i; j++)
        {
            hash[sg[j]^sg[i-j]] = 1;
        }
        int j;
        for(j=0; j<MAXN; j++)
            if(!hash[j])
                break;
        sg[i] = j;
    }
}
int main()
{
    Get_sg();
    int T;
    scanf("%d",&T);
    for(int cas=1; cas<=T; cas++)
    {
        int m, sum = 0;
        scanf("%d",&m);
        for(int i=0; i<m; i++)
        {
            int x;
            scanf("%d",&x);
            sum ^= sg[x];
        }
        if(sum)
            printf("Case %d: Alice\n",cas);
        else
            printf("Case %d: Bob\n",cas);
    }
    return 0;
}
 

LIGHT OJ 1199 - Partitioning Game