UESTC - 1519 Picking&Dancing

Yitong_Qin and Xiaoyu_Chen are playing a game.There are N stones placed on the ground,forming a sequence.Thr stones are labeled from 1 to N.Yitong_Qin and Xiaoyu_Chen in turns take exactly two consecutive stones on the ground until there are no consecutive stones on the ground.That is,each player can take stone i and stone i+1 ,where 1≤i≤n−1

.If the number of stones left is odd,Yitong_Qin wins ,so Xiaoyu_Chen has to dance.Otherwise Xiaoyu_Chen wins Yitong_Qin has to dance.

Input

The input contains an integer N(1≤N≤5000)

,the number of stones.

Output

Output the guy has to dance.

Sample input and output

Sample Input	Sample Output
1	Xiaoyu_Chen
2	Yitong_Qin
5	Xiaoyu_Chen

题意：Yitong_Qin跟Xiaoyu_Chen两个人轮流取石子，石子共有N个，最后剩下的是奇数，Yitong_Qin赢， 所以Xiaoyu_Chen跳舞，否则Yitong_Qin跳舞，输出要跳舞的人。
解题思路：直接看N是奇数还是偶数。

代码：

#include<cstdio>
int main(){
	int N,mod;
	scanf("%d",&N);
	mod=N%2;
	if(mod==1){
		printf("Xiaoyu_Chen");
	}
	else{
		printf("Yitong_Qin");
	}
	return 0;
}