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时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述
kiki和bob在玩一个游戏，他们收集了一些石子，约定两人轮流从这些石子中取出一部分，但是每次只能取1、2或3颗石子，无法继续取的人输了，kiki和bob都非常聪明，他们总是按照最有利于自己的方式进行游戏，请你预测最后谁会赢得游戏，如果先手胜利输出kiki，否则输出bob。
输入描述:
输入一个整数n（1<=n<=10^9）代表一共有多少颗石子。
输出描述:
如果先手胜利输出kiki，否则输出bob。
示例1
输入
10
输出
kiki
示例2
输入
8
输出
bob
思路：
可以先列举几个数试一下：
1、若只有1 、2、3个石子，先手必赢；
2、若有4个石子，先手必输；
3、若有5、6、7个石子，先手可让石子的状态变成4个，则后手面对4个石子的局面必输；
4.若有8个石子，先手会让后手面对第三种情况，则自己必输，后面以此类推；
综上可以得出结论：若有（（3 + 1）* n）个石子，先手必输；
代码：
#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
using namespace std;
int main()
{
int n;
cin >> n;
if(n % 4 == 0)
{
cout << "bob" << endl;
}
else
{
cout << "kiki" << endl;
}
return 0;
}
