http://codeforces.com/problemset/problem/869/A
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
input
3
1 2 3
4 5 6
output
Karen
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
有两组数据，两组数据之间两两异或，若异或后的数仍在数组中则ans++，若最后ans%2==0，则karen赢，否则koyomi赢。
题解
貌似A题都不会怎么卡大数据时间啥的
直接暴力求解
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <functional>
#include <time.h>
using namespace std;
int vis[10000000],a[2010],b[2010];
int main()
{
memset(vis,0,sizeof(vis));
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
vis[a[i]]=1;
}
for(int i=0;i<n;i++)
{
scanf("%d",&b[i]);
vis[b[i]]=1;
}
int sum=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
if(vis[a[i]^b[j]])
sum++;
if(sum%2==0)
cout << "Karen" << endl;
else
cout << "Koyomi" << endl;
}
更简单的写法：
将两个数组分为a数组和b数组，有ai^bj=c,如果c属于b，则存在ai^c=bj，如果c属于a，则存在bj^c=ai，如果不存在，那就是0，对结果无影响。也就是说符合情况的两个数异或然后结果肯定是成对出现的，所以答案一定是 Karen
#include <stdio.h>
int main()
{
printf("Karen\n");
}