POJ 2348-Euclid's Game(博弈论)

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Euclid's Game

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2348

Appoint description:

Description

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 1215 240 0

Sample Output

Stan winsOllie wins

题意：有两个数a,b，用大数减去小数的倍数，先有一个变成0的那个人获胜。

思路：有三种情况(假设a为大数）：

1，如果a%b==0，那肯定是先手赢。

2，如果a<2*b，那么下一步只能变成(a-b,b)，那么赢得几率是相互交替的。

3，如果a>2*b,那么肯定是先手赢，因为总可以转变成(a，a%b)和(a，a%b+b）的状态，因为这两种状态和第二种情况是相邻的，所以肯定有一种处于必败状态，先手只要找到这种状态即可。

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace std;typedef long long LL;int main(){    LL a,b;    int flag=0;//对于第二种交替情况，看谁赢。    while(~scanf("%lld %lld",&a,&b)){        if(!a&&!b) break;        flag=0;        for(;;){            if(a<b) swap(a,b);            if((a%b==0)||(a>2*b)) break;            a-=b;            flag=1-flag;        }        if(!flag)            printf("Stan wins\n");        else            printf("Ollie wins\n");    }    return 0;}

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