SDNU 1499.Problem

1499.Problem_H Time Limit: 1000 MS    Memory Limit: 32768 KB Description Spring is coming, too.(Why do I say too?) SDNU is a school what has a gender imbalance. There are some beautiful girls in the SDNU ACM-ICPC Association but HuaHua(HH) and ChaoChao(CC) love a little fat girls. Fortunately, there is a girl named Virgo. HH and CC love her and even fight for her. And then one day, virgo comes up a solution about this problem. Virgo stands at the end of a queue. The queue contains N people. Let HH and CC start from the head of the queue, and each one every time only can choose to take away 1-4 people. When someone can take Vrigo, Vrigo will give him a very beautiful night.

(HH take away people at first and everyone will choose the most beneficial strategy for himself.)

Input The first line is a number N(N<=500), represents N test case.

Then followed N lines,each line have a number(<=2^1000) mesns the number of the queue.

Output

Tell us who will have a beautiful night.(“huahua” or “chaochao”)

Sample Input 3 2 3 6 Sample Output huahua huahua huahua

    一道典型的巴什博奕，简单说就是给出要找的人的位置，HH和CC一次可以选择往前找1~4个人，HH先找，问谁先找到，假设他们都选择最佳的方案。设n为人数，m为一次找的最多人数，则n=m+1时，由于一次最多只能取m个，所以无论先取者拿走多少个，后取者都能够一次拿走剩余的物品，后者取胜，所以当一方面对的局势是n%(m+1)=0时，其面临的是必败的局势。所以可以判断当（n-1）%（m+1）==0时后手胜利。

    而这道题特殊的地方在于这道题的数据，给出的位置的范围最大是2^1000，所以这道题不能直接long long处理，而要采用大数的方法来进行处理了，代码是队友写的Java代码，这里copy过来学习一下Java大数的写法→_→

    下面是AC代码：

import java.math.*; import java.util.Scanner; public class Main { public static void main(String[] args) throws Exception { Scanner cin=new Scanner(System.in); int n; while(cin.hasNext()) { n=cin.nextInt(); for(int i=0;i<n;i++) { BigDecimal a=cin.nextBigDecimal(); BigDecimal b=new BigDecimal(5D); BigDecimal c=new BigDecimal(0D); BigDecimal num=a.remainder(b); if(num.compareTo(c) == 0) System.out.println("chaochao"); else System.out.println("huahua"); } } } }