srm 558 backup 1 DIV 2

Problem Statement
	Cat and Rabbit are going to play the following game. There are some tiles in a row. Each of the tiles is colored white or black. You are given a stringtiles, describing the initial arrangement of tiles. The characters '.' and '#' represent a white tile and a black tile, respectively. Cat and Rabbit take alternating turns. Cat plays first. In each turn, the following actions must be performed: First, the player must select a black tile and step on it. Then, the player must make some steps (as many as they want, but at least one). In each step, the player must select an adjacent white tile, move to that tile, and change its color to black. The animal who is unable to take a valid turn loses the game. Return the string "Cat" if Cat will win, "Rabbit" if Rabbit will win, assuming both animals play optimally.
Definition
	Class: CatAndRabbit Method: getWinner Parameters: string Returns: string Method signature: string getWinner(string tiles) (be sure your method is public)
Constraints
-	tiles will contain between 1 and 50 characters, inclusive.
-	Each character in tiles will be '.' or '#'.
Examples
0)
	"#.." Returns: "Cat" In the first turn, Cat will select the leftmost tile and can move to the right twice. Then Rabbit has no legal move and loses.
1)
	".#." Returns: "Rabbit" In the first turn, Cat must select the middle tile. Then, Cat can decide whether to step to the left or to the right. In the second turn, Rabbit will select the middle tile and then step in the opposite direction.
2)
	"###" Returns: "Rabbit" Cat has no legal move at the beginning of the game.
3)
	"#..##.#" Returns: "Cat"
4)
	"..." Returns: "Rabbit"
5)
	"###...####....###...####...." Returns: "Rabbit"

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.     

#include<math.h>
#include<vector>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>




using namespace std;


#define FOR(I,A,B) for (int I=int(A);I<int(B);I++)
#define PUSH(X,N,n) {FOR(q,0,N){X.push_back(n);}}
#define  ll _int64


template<class T>  T max (T A, T B){if(A>B) return A; else return B;} 
template<class T>  T min (T A, T B){if(A<B) return A; else return B;} 


//#define ll long long
//typedef  __int64 ll


#define oo (1<<25)
#define SZ(x) (int)x.size()






#define MAX 100


string rabbit="Rabbit";
string cat="Cat";


class CatAndRabbit{
public:
string getWinner(string tiles)
{
int ok=0;
for(int q=0;q<SZ(tiles);q++)
if(tiles[q]=='#') ok=1;
if(!ok) return rabbit;

int i,j,ans=0;
for( i=0;i<SZ(tiles);)
{
if(tiles[i]=='#') i++;
else if(tiles[i]=='.')
{
int len=1;
for(int j=i+1;j<SZ(tiles);j++)
{
if(tiles[j]=='.') len++;
else break;
}
i=j;
if(len) ans^=len;
}
}
if(ans) return cat;
else 
return rabbit;
}
};

前面讲的错了，看了别人的我突然发现自己弱了。。。。。。。

比如一个数据 2 4  6   好像很难发现到底谁先走完。

后来我想到吧他分解  2 2+2  2+4  可是这样并没什么用。因为 你分解成2+2 别人去3 就没办法了。

可是如果分解成2进制 0010   0100 0110 就可以发现。无论怎么改变其中的一个数，总是可以出现互补的。（就是各个位都有偶数个1）

如 第一步 0001 0100 0110

那么第2步很容易想到  0001 01000101

也就是无论第一个怎么走他都摆脱不了输的命啊