HDU 3389 Game(博弈)

Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 214    Accepted Submission(s): 150

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

 

 

 

想法太神了！！

 

1. 分成一个二分图
如果可以从A拿卡片到B，连一条从A到B的边。把所有box编号x满足（(x%


3==0&&x%2==1) || x%3==1)这个条件的放左边，其他放右边，不难发现
a) 只有从左边到右边的边或从右到左的边。
b) 所有不能拿卡片出去的box都在左边。
2. 证明左边的box并不影响结果。假设当前从右边的局势来看属于输家的人为了
摆脱这种局面，从左边的某盒子A拿了n张卡片到B，因为B肯定有出去的边，对手
会从B再取走那n张卡片到左边，局面没有变化
3. 于是这就相当于所有右边的box在nim游戏。

 

 

其实只有1 3 4 是不能移动的。

 

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
int main()
{
    int T;
    int n;
    int a;
    scanf("%d",&T);
    int iCase = 0;
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        int sum = 0;
        for(int i = 1;i <= n;i++)
        {
            scanf("%d",&a);
            if((i%3==0 && i%2==0) || i%3 == 2)
                sum ^= a;
        }
        if(sum)printf("Case %d: Alice\n",iCase);
        else printf("Case %d: Bob\n",iCase);
    }
    return 0;
}