题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6312
Problem Description
Alice and Bob are playing a game.
The game is played on a set of positive integers from 1 to n.
In one step, the player can choose a positive integer from the set, and erase all of its divisors from the set. If a divisor doesn't exist it will be ignored.
Alice and Bob choose in turn, the one who cannot choose (current set is empty) loses.
Alice goes first, she wanna know whether she can win. Please judge by outputing 'Yes' or 'No'.
The game is played on a set of positive integers from 1 to n.
In one step, the player can choose a positive integer from the set, and erase all of its divisors from the set. If a divisor doesn't exist it will be ignored.
Alice and Bob choose in turn, the one who cannot choose (current set is empty) loses.
Alice goes first, she wanna know whether she can win. Please judge by outputing 'Yes' or 'No'.
Input
There might be multiple test cases, no more than 10. You need to read till the end of input.
For each test case, a line containing an integer n. (1≤n≤500)
For each test case, a line containing an integer n. (1≤n≤500)
Output
A line for each test case, 'Yes' or 'No'.
Sample Input
1
Sample Output
Yes
题意：
A和B玩游戏，有一个1~n的数字序列，规定每人每次能从里面挑出一个数，删去它以及他的因数，A先手，A想知道自己能不能赢。
题解：
假设现在是在2~n的序列里玩游戏，那么
如果A先手情况下A是胜者，那么A就这么玩，因为1是任何数的因数，A还是能赢；
如果A先手情况下B是胜者，那么A就先拿掉一个1，那么就变成2~n的序列下B先手，A就依然是胜者；
综上，先手必胜。
AC代码：
#include<bits/stdc++.h> using namespace std; int main(){for(int n;cin>>n;) cout<<"Yes"<<endl;}