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

A. The Artful Expedient

#include<bits/stdc++.h>
using namespace std;
const int maxn = 20000;
const int maxn2 = 2e7+100;
int num1[maxn],num2[maxn];
bool num[maxn2];
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&num1[i]);
num[num1[i]] = true;//标记ai中出现过的数
}
for(int i=0;i<n;i++)
{
scanf("%d",&num2[i]);
num[num2[i]] = true;//标记bi中出现过的数
}

//数据量比较小直接暴力双重循环
int sum = 0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
int now = num1[i]^num2[j];
if(num[now])
sum++;
}

if(sum%2)
printf("Koyomi");
else
printf("Karen");

return 0;
}