Codeforces--896A--The Artful Expedient

题目描述：
Rock… Paper!

After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.

A positive integer n is decided first. Both Koyomi and Karen independently choose n distinct positive integers, denoted by x1, x2, …, xn and y1, y2, …, yn respectively. They reveal their sequences, and repeat until all of 2n integers become distinct, which is the only final state to be kept and considered.

Then they count the number of ordered pairs (i, j) (1 ≤ i, j ≤ n) such that the value xi xor yj equals to one of the 2n integers. Here xor means the bitwise exclusive or operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.

Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you’re here to help determine the winner of their latest game.
输入描述:
The first line of input contains a positive integer n (1 ≤ n ≤ 2 000) — the length of both sequences.

The second line contains n space-separated integers x1, x2, …, xn (1 ≤ xi ≤ 2·106) — the integers finally chosen by Koyomi.

The third line contains n space-separated integers y1, y2, …, yn (1 ≤ yi ≤ 2·106) — the integers finally chosen by Karen.

Input guarantees that the given 2n integers are pairwise distinct, that is, no pair (i, j) (1 ≤ i, j ≤ n) exists such that one of the following holds: xi = yj; i ≠ j and xi = xj; i ≠ j and yi = yj.
输出描述:
Output one line — the name of the winner, that is, “Koyomi” or “Karen” (without quotes). Please be aware of the capitalization.
输入：
3
1 2 3
4 5 6
5
2 4 6 8 10
9 7 5 3 1
输出：
Karen
Karen
题意：
上面的数和下一个数两两组合, 每一个组合中如果异或后的值在输入中出现过.ans++,ans为偶数Karen赢,反之另一位赢…
题解：
a^b=c;
a^ c = a^ b ^a=b;
所以如果x与y中分别取一个数a和b求异或值c，如果c在y中，那a^c=b；如果c在x中，那b ^ c=a;
也就是无论如何，答案一定是偶数。
代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
using namespace std;const int maxn = 2000000 + 5;
int a[maxn],b[maxn];int main(){int n,x;while(scanf("%d",&n)!=EOF){map<int,int> mp;//memset(vis,0,sizeof(vis));for(int i = 0; i < n; i ++) scanf("%d",&a[i]);for(int i = 0; i < n; i ++) scanf("%d",&b[i]);printf("Karen\n");}return 0;
}

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