Day11 - H - Euclid's Game HDU - 1525

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.

InputThe input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts.OutputFor 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

思路：一开始考虑PN状态是否跟奇偶有关，举出反例后pass，那么就找规律，x 0的状态是P状态，可以确定N状态为 x y (y%x=0)，那么当y>=2*x时就可以判断状态，y==2*x时，一定是N状态，y>2*x时，先手可以判断y%x x是否为必败态，若y%x x为必败态，先手就将该状态转移给后手，否则，先手就将y%x+x x转移给后手，后手一定只能把y%x x转移给先手，先手必胜，也就是说，当y>=2*x时，先手就掌握了比赛，是必胜态

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;


void run_case() {
    int n, m;
    while(cin >> n >> m && n + m) {
        int win = 1;
        if(n < m) swap(n, m);
        while(m) {
            if(n%m == 0 || (n / m) > 1) break;
            n -= m;
            swap(n, m);
            win ^= 1;
        }
        if(win) cout << "Stan wins\n";
        else cout << "Ollie wins\n";
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    //int t; cin >> t;
    //while(t--)
    run_case();
    cout.flush();
    return 0;
}

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