A. The Artful Expedient

Problem Statement

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.

Input

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.

Output

Output one line — the name of the winner, that is, “Koyomi” or “Karen” (without quotes). Please be aware of the capitalization.

Examples

Example 1
    Input
        3
        1 2 3
        4 5 6
    Output
        Karen
Example 2
    Input
        5
        2 4 6 8 10
        9 7 5 3 1
    Output
        Karen

Note

In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again.

题意

有两个人在玩游戏，每个人有一段长为n的序列，且这2*n个数全都不同。问你其中有几对数对满足从第一段序列中选出一个数和第二段序列中选出一个数，他们的异或和在这2*n个数中存在。如果有偶数对，那么Karen赢，否则Koyomi赢。

思路

最暴力的方法，当然是直接n^2*log(n)暴力扫然后用set存这2n个数啦；
其实还有另一种解法，因为如果ai xor bj = aka_i\ xor\ b_j\ =\ a_k，那么ak xor bj = aia_k\ xor\ b_j\ =\ a_i。所以肯定都是偶数对偶数对出现的，只要直接puts(“Karen”);就行啦！

Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline void readInt(int &x) {x=0;int f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();x*=f;
}
inline void readLong(ll &x) {x=0;int f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();x*=f;
}
/*================Header Template==============*/
int a[2005],n,x[2005],ans=0;
set<int> sa,sb;
int main() {readInt(n);for(int i=1;i<=n;i++)readInt(a[i]),sa.insert(a[i]);for(int i=1;i<=n;i++)readInt(x[i]),sb.insert(x[i]);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(sa.count(a[i]^x[j])||sb.count(a[i]^x[j]))ans++;
//  cout<<ans<<endl;if(ans&1)puts("Koyomi");elseputs("Karen");return 0;
}

Another Code

#include<cstdio>
main(){puts("Karen");}

在PHP语言里，只需要5B代码：Karen，你就能A掉此题啦！

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Codeforces Round #439 (Div. 2) A. The Artful Expedient

A. The Artful Expedient

Problem Statement

Input

Output

Examples

Note

题意

思路

Code

Another Code

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