HDU - 1525(31/600)

最新推荐文章于 2020-08-04 09:23:04 发布

原创最新推荐文章于 2020-08-04 09:23:04 发布 · 264 阅读

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ACM练习同时被 2 个专栏收录

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本文介绍了一种基于两个自然数的简单博弈游戏，通过玩家之间的轮换操作来决定胜负，并提供了一个C++实现示例。游戏的核心在于如何通过数学运算控制游戏进程，从而赢得比赛。

Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtracts any positive multiple of the lesser of the two numbers from the greater of the two numbers, provided that the resulting number must be nonnegative. Then Ollie, the second player, does the same with the two resulting numbers, then Stan, etc., alternately, until one player is able to subtract a multiple of the lesser number from the greater to reach 0, and thereby wins. For example, the players may start with (25,7):

25 7
11 7
4 7
4 3
1 3
1 0

an Stan wins.

Input
The input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts.
Output
For each line of input, output one line saying either Stan wins or Ollie wins assuming that both of them play perfectly. The last line of input contains two zeroes and should not be processed.

Sample Input
34 12
15 24
0 0
Sample Output
Stan wins
Ollie wins

博弈的过程就是一只不停的互相减
关键在于超过2倍的时候
他可以控制胜败…

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    int n, m;
    while (cin >> n >> m)
    {
        if (n == 0 && m == 0)break;
        int da = max(n, m), xiao =min(n, m);
        int daan=0;
        for (int a = 1;; a++)
        {
            if (xiao == 0||da>=2*xiao||da%xiao==0)break;
            int cs = da / xiao;
            da -= cs*xiao;
            swap(da, xiao);
            daan ++;
        }
        if (daan % 2==0)printf("Stan wins\n");
        else printf("Ollie wins\n");
    }
}