Game

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 50 Accepted Submission(s): 44

Problem Description Bob and Alice are playing a new game. There are n boxes which have been numbered from 1 to n. Each box is either empty or contains several cards. Bob and Alice move the cards in turn. In each turn the corresponding player should choose a non-empty box A and choose another box B that B<A && (A+B)%2=1 && (A+B)%3=0. Then, take an arbitrary number (but not zero) of cards from box A to box B. The last one who can do a legal move wins. Alice is the first player. Please predict who will win the game.

Input The first line contains an integer T (T<=100) indicating the number of test cases. The first line of each test case contains an integer n (1<=n<=10000). The second line has n integers which will not be bigger than 100. The i-th integer indicates the number of cards in the i-th box.

Output For each test case, print the case number and the winner's name in a single line. Follow the format of the sample output.

Sample Input 2 2 1 2 7 1 3 3 2 2 1 2

Sample Output Case 1: Alice Case 2: Bob

Author hanshuai@whu

Source The 5th Guangting Cup Central China Invitational Programming Contest

Recommend notonlysuccess

/*
題意：1到n的盒子裏，是空的或者裝着卡片，每人輪流移動卡片，Alice先手，移動卡片的規則：每次選擇一個A，而後選擇一個B
    要求 B<A&&(A+B)%2==1&&(A+B)%3==0,從A中抽取任意一張卡片放到B中，誰不能移動了，就輸了。

初步思路：這個就是先將能移動的盒子對找出來，而後統計能移動的牌

#錯誤：上面的想法不完善，只是一方面的認爲卡片會從一上一級向下一級移動。

#補充：階梯博弈，實際上卡片是在3的餘數上進行轉移的。

#感悟：手殘用了數組，開小了，wa了兩發才發現
*/
#include<bits/stdc++.h>
using namespace std;
int n;
int x;
int res=0;
int t;
void init(){
    res=0;
}
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++){
        init();
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            if(i%6==0||i%6==2||i%6==5)
                res^=x;
        }
        if(res) printf("Case %d: Alice\n",ca);
        else printf("Case %d: Bob\n",ca);
    }
    return 0;
}