2016 ICPC大连赛区 hdu5978 To begin or not to begin

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探讨了在一个包含k个黑球和1个红球的游戏里，先手玩家与后手玩家赢得游戏的概率分析。通过数学概率计算得出，当k为奇数时，双方胜率相等；k为偶数时，先手玩家拥有更高胜率。

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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): 229 Accepted Submission(s): 175

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：给k个黑球，1个红球，谁先拿到红球谁就赢，问先手赢的几率大还是后手赢的几率大或者一样大。

思路：这是个简单的概率问题，第n回合抽到红球的几率一定等于1/（k+1），那么赢的几率=拥有的回合数*（1/（k+1）），如果k为奇数，那么两边回合一样所以几率一样大，如果k为偶数那么必定先手的有多一个回合的机会，所以几率更大，下面给代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
typedef long long LL;
using namespace std;
#define maxn 1005
#define ll l,mid,now<<1
#define rr mid+1,r,now<<1|1
#define lson l1,mid,l2,r2,now<<1
#define rson mid+1,r1,l2,r2,now<<1|1
int main(){
    int k;
    while(~scanf("%d",&k)){
        if(k&1)
            printf("0\n");
        else
            printf("1\n");
    }
}