HDU 5978 To begin or not to begin（概率）

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探讨了在一个包含多个黑球和一个红球的游戏场景中，两位玩家轮流抽取，先抽到红球者获胜。分析了先手与后手的优势，并给出了一种简单的判断方法。

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To begin or not to begin

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1226 Accepted Submission(s): 790

Problem Description

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

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

题意：

n个黑球，1个红球，两个人依次摸，摸到红球算赢。问你先摸好还是后摸好，还是都一样。

POINT：

可知当n是奇数时，两个人摸的机会一样。输出0

n偶数时，先摸的可以多摸一次，输出1。

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
const double PI = acos(-1.0);
int main()
{
    int k;
    while(~scanf("%d",&k)){
        if(k&1) printf("0\n");
        else printf("1\n");
    }
    return 0;
}