HDU 3389 Game（nim博弈）

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 54 Accepted Submission(s): 48 

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

221 271 3 3 2 2 1 2

 

Sample Output

Case 1: AliceCase 2: Bob

 

Author

hanshuai@whu

 

Source

The 5th Guangting Cup Central China Invitational Programming Contest

 

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notonlysuccess

 

//1、3、4点不可再移动，1、3、4点为终点。//打表发现规律，x % 6 == 2 || 5 || 0 的数到达1、3、4点一定是通过奇数步。//其他的点，一定要先走到奇数步的点，才能到达1、3、4点。//偶数步的点不会影响结果//这就好像nim博弈取石子那样了，所以对奇数步的点，进行异或求解就好了#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#include <queue>using namespace std;int main() {    int t;    cin >> t;    for (int i = 1; i <= t; ++i) {        int ans = 0;        int n;        cin >> n;        for (int j = 1; j <= n; ++j) {            int a;            cin >> a;            if (j % 6 == 2 || j % 6 == 5 || j % 6 == 0) {                ans ^= a;            }        }        if (ans) {            cout << "Case " << i << ": Alice\n";        } else {            cout << "Case " << i << ": Bob\n";        }            }}