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.InputMultiple 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.
OutputFor 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

在每一次概率相同的情况下，拿的次数多的人赢的概率大。至于每一次赢概率为什么相同，就跟概率论里的买彩票一样，先买后买概率一样。
 1 #include <iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<string>
 6 #include<queue>
 7 #include<vector>
 8 #include<cmath>
 9 using namespace std;
10 const double pi=acos(-1.0);
11 int n,d,x;
12 int main()
13 {
14      while(~scanf("%d",&n))
15     {
16       if (n%2==0) printf("1\n");
17         else printf("0\n");
18     }
19     return 0;
20 }