hdu 1851（A Simple Game）（sg博弈）

A Simple Game

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1487    Accepted Submission(s): 939

Problem Description

Agrael likes play a simple game with his friend Animal during the classes. In this Game there are n piles of stones numbered from 1 to n, the 1st pile has M₁ stones, the 2nd pile has M₂ stones, ... and the n-th pile contain M_n stones. Agrael and Animal take turns to move and in each move each of the players can take at most L₁stones from the 1st pile or take at most L₂ stones from the 2nd pile or ... or take L_n stones from the n-th pile. The player who takes the last stone wins.

After Agrael and Animal have played the game for months, the teacher finally got angry and decided to punish them. But when he knows the rule of the game, he is so interested in this game that he asks Agrael to play the game with him and if Agrael wins, he won't be punished, can Agrael win the game if the teacher and Agrael both take the best move in their turn?

The teacher always moves first(-_-), and in each turn a player must takes at least 1 stones and they can't take stones from more than one piles.

 

Input

The first line contains the number of test cases. Each test cases begin with the number n (n ≤ 10), represent there are n piles. Then there are n lines follows, the i-th line contains two numbers M_i and L_i (20 ≥ M_i > 0, 20 ≥ L_i > 0). 

 

Output

Your program output one line per case, if Agrael can win the game print "Yes", else print "No". 

 

Sample Input

2 1 5 4 2 1 1 2 2

 

Sample Output

Yes No

 

Author

Agreal@TJU

 

Source

HDU 2007 Programming Contest - Final

 

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简单的SG函数题，结果为每个子游戏的sg值的异或，每个子游戏都是简单的巴什博弈

#include<stdio.h>
int main()
{
    int t,n,m,l;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int sg=0;
        for(int i=0;i<n;i++){
            scanf("%d%d",&m,&l);
            sg ^= m%(l+1);
        }
        if(sg)
            puts("No");
        else
            puts("Yes");
    }
    return 0;
}