POJ 2488-A Funny Game(简略博弈)

POJ 2488-A Funny Game(简单博弈)

A Funny Game

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2484

Appoint description:

Description

Alice and Bob decide to play a funny game. At the beginning of the game they pick n(1 <= n <= 10 ⁶) coins in a circle, as Figure 1 shows. A move consists in removing one or two adjacent coins, leaving all other coins untouched. At least one coin must be removed. Players alternate moves with Alice starting. The player that removes the last coin wins. (The last player to move wins. If you can't move, you lose.)
POJ 2488-A Funny Game(简略博弈)

Figure 1
Note: For n > 3, we use c1, c2, ..., cn to denote the coins clockwise and if Alice remove c2, then c1 and c3 are NOT adjacent! (Because there is an empty place between c1 and c3.)

Suppose that both Alice and Bob do their best in the game.
You are to write a program to determine who will finally win the game.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (1 <= n <= 10 ⁶). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

For each test case, if Alice win the game,output "Alice", otherwise output "Bob".

Sample Input

Sample Output

Alice
Alice
Bob

题意：给出N个硬币围成一个圈，然后两个人从这圈硬币中轮流拿1个或毗邻的2个硬币。直到全部拿完为止，最后一个拿的人为，胜者。

思路：慢慢的把列举几种情况就可以发现，当n<3的时候先手赢，当n>=3的时候后手赢

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
int main()
{
    long long n;
    while(~scanf("%lld",&n)){
        if(!n) break;
        if(n<3)
            printf("Alice\n");
        else
            printf("Bob\n");
    }
    return 0;
}

POJ 2488-A Funny Game(简略博弈)

相关推荐