hdu3404 Switch lights

Switch lights

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 341 Accepted Submission(s): 180

Problem Description

lxhgww is playing a game with his computer Deep Blue.
The game is played on a matrix containing lights. At first, some lights are on, while others are off. lxhgww and Deep Blue take turns to switch the lights. For each step, the player should choose a rectangle in the matrix: (x1 , y1) , (x1 , y2) , (x2 , y1) , (x2 , y2) , (x1<=x2,y1<=y2, the light at (x2, y2) should be on) and change the lights’ status on the four vertex of the rectangle, namely on to off, and off to on. The player turns all the lights off wins the game. Notice the rectangle is possibly degenerated to line or even a single cell so that the player may also switch two or one besides four lights in a move.
Deep Blue's strategy is perfect, if it has a chance to win, never will it lose. Does lxhgww have a chance to win if he takes the first step?

Input

The first line is an integer T(T<=100) indicating the case number.
Each case has one integers n (n<= 1000 ), the number of on-lights at the beginning of the game.
Then come n lines, each line has two integers, xi , yi, (1<=xi<=10000, 1<=yi<=10000) , so light at (xi, yi) is on at first. (No two lights at the same position)

Output

If lxhgww still has a chance to win, output "Have a try, lxhgww.", otherwise tell lxhgww "Don't waste your time."

Sample Input

   
   
   
   

    
    
    
    2
2
1 2
2 1
2
1 1
2 2
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    Don't waste your time.
Have a try, lxhgww.
   
   
   
   
   
   
   
   

    
    
    
    坑啊,这题不好做,推荐看从“k 倍动态减法游戏”出发探究一类组合游戏问题这篇论文,很详细,但是不好懂!
   
   
   
   
   
   
   
   

    
    
    
    
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int pri[2][2]={0,0,0,1};
int get(int x,int y){
    if(x<2)return pri[x][y];
    int i,p,t,s,m;
    for(i=0;;i++){
        if(((1<<(1<<i))<=x)&&(x<(1<<(1<<(i+1)))))
        break;
    }
    m=1<<(1<<i);
    p=x/m,s=y/m,t=y-s*m;
    int c1=get(p,s);
    int c2=get(p,t);
    return m*(c1^c2)^get(m/2,c1);

}
int getsg(int x,int y){
    if(x<y)return getsg(y,x);
    if(x<2)return pri[x][y];
    int i,p,t,s,q,m;
    for(i=0;;i++){
        if(((1<<(1<<i))<=x)&&(x<(1<<(1<<(i+1)))))
        break;
    }
    m=1<<(1<<i);
    p=x/m,q=x-p*m,s=y/m,t=y-s*m;
    int c1=getsg(p,s);
    int c2=getsg(p,t)^getsg(q,s);
    int c3=getsg(q,t);
    return m*(c1^c2)^c3^get(m/2,c1);
}
int main()
{
    int tcase,i,x,y,ans,n;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%d",&n);
        for( i=0,ans=0;i<n;i++){
            scanf("%d%d",&x,&y);
            ans^=getsg(x,y);
        }
        if(ans)printf("Have a try, lxhgww.\n");
        else printf("Don't waste your time.\n");
    }
    return 0;
}