Euclid's Game

 

Euclid's Game

 1000(ms)

 65535(kb)

 720 / 2828

Starts with two unequal positive numbers (M,N and M>N) on the board. Two players move in turn. On each move, a player has to write on the board a positive number equal to the difference of two numbers already on the board; this number must be new, i.e., different from all the numbers already on the board. The player who cannot move loses the game. Should you choose to move first or second in this game? 

According to the above rules, there are two players play tihs game. Assumptions A write a number on the board at first, then B write it. 

Your task is write a program to judge the winner is A or B.

输入

Two unequal positive numbers M and N , M>N (M<1000000)

输出

A or B

样例输入

3 1

样例输出

A

 

经过我terrible的英语，终于明白了这道题就是关于欧几里得算法的题目，大概意思就是：

棋盘上以两个不等的正数（M，N和M> N）开始。两名球员依次移动。在每次移动中，玩家必须在棋盘上写上一个正数，该数字等于棋盘上已有的两个数字的差值; 这个数字必须是新的，即与主板上已有的所有数字不同。无法移动的玩家输掉游戏。你应该选择在这场比赛中先手还是后手？
根据以上规则，有两名玩家玩游戏。假设A首先在板上写一个数字，然后B写它。 
你的任务是写一个程序来判断胜利者是A还是B.

每次写的数，只能是最开始两个数的最大公约数的p（p最大为M/gcd（M,N））倍，所以判断M/gcd（M,N）奇偶性即可。

#include
using namespace std;
//欧几里得最大公约数 
int gcd(int a, int b) {
	if(b==0) return a;
	return gcd(b,a%b);
}

int main() {
	int m,n,p;
	cin >> m;
	cin >> n;
	p = gcd(m,n);
	//求得最大公约数，如果该A玩家取数/除以最大公约数余2不等于0，则表明该玩家能赢 
	if(( m / p) % 2 != 0)
        cout << "A" << endl;
	else
        cout << "B" << endl;
	return 0;
}