2016ACM/ICPC亚洲区大连站 H - To begin or not to begin

最新推荐文章于 2021-07-14 22:14:09 发布

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本文探讨了一个简单的博弈论游戏，其中两名玩家从装有黑色和红色球的盒子中轮流抽取球，直到抽到红球为止。文章提供了一段C++代码，用于判断在不同数量的黑球情况下，先手玩家是否有优势。

A box contains black balls and a single red ball. Alice and Bob draw balls from this box without replacement, alternating after each draws until the red ball is drawn. The game is won by the player who happens to draw the single red ball. Bob is a gentleman and offers Alice the choice of whether she wants to start or not. Alice has a hunch that she might be better off if she starts; after all, she might succeed in the first draw. On the other hand, if her first draw yields a black ball, then Bob’s chances to draw the red ball in his first draw are increased, because then one black ball is already removed from the box. How should Alice decide in order to maximize her probability of winning? Help Alice with decision.

Input
Multiple test cases (number of test cases≤50), process till end of input.
For each case, a positive integer k (1≤k≤10^5) is given on a single line.

Output
For each case, output:
1, if the player who starts drawing has an advantage
2, if the player who starts drawing has a disadvantage
0, if Alice’s and Bob’s chances are equal, no matter who starts drawing
on a single line.

Sample Input
1
2

Sample Output
0
1

最初有一点题意理解错了，输入的是黑球数而不是总共的球数，剩下的就是找规律了，黑球数为奇数时平手，为偶数时先手优势。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<math.h>

using namespace std;

int main(void){
    int num;

    while(scanf("%d",&num) != EOF){
        if(num %2 == 0)
            printf("1\n");
        else
            printf("0\n");
    }


    return 0;
}