Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 184 Accepted Submission(s): 132
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
Source
Recommend
题意：给你一串数字，A和B分别可以对这串数字进行操作，每次操作可以删去一个数及他所有的因子，轮到谁时谁无法再进行操作谁就输了
首先考虑1-n，我们可以先不看1，则2-n一定有一个胜者，如果这个状态A胜的话，我们就可以选择那个让我们胜的数，因为1是所有数的因子所以1也会被删去，如果A输，我们可以先去掉1来转变状态。这样A还是会胜。
综述A一定会获胜
#include<iostream>
using namespace std;
int main() {
int n;
while( cin >> n ) {
cout << "Yes" << endl;
}
return 0;
}