Euclid's Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1116 Accepted Submission(s): 516
Problem 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.
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
Source
Recommend
LL
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1525
题意：给出两个正数。
两个人轮流进行操作：
较大的数减掉较小的数的倍数，得到两个非负数。
谁先先把其中一个数减为0的获胜。问谁可以赢。Stan是先手。
假设两个数为a,b（a>=b)
如果a==b.那么肯定是先手获胜。一步就可以减为0,b
如果a%b==0.就是a是b的倍数，那么也是先手获胜。
如果a>=2*b. 那么 那个人肯定知道a%b,b是必胜态还是必败态。如果是必败态，先手将a,b变成a%b,b,那么先手肯定赢。如果是必胜态，先手将a,b变成a%b+b,b.那么对手只有将这两个数变成a%b,b,先手获胜。
如果是b<a<2*b 那么只有一条路：变成a-b,b (这个时候0<a-b<b).这样一直下去看谁先面对上面的必胜状态。
/* * 如果面对a==b 或者a>=2*b则先手比赢 * 如果b<a<2*b.那么只有一条路，变成a-b,b; * 这个时候互换大小变成b,a-b. * 这样一直循环，看谁面对a==b || a>=2*b的情况 */ #include <iostream> #include <string.h> #include <algorithm> #include <stdio.h> using namespace std; int main() { int a,b; while(scanf("%d%d",&a,&b)==2) { if(a==0||b==0)break; if(a<b)swap(a,b); bool flag=true; while(1) { if(a==b || a/b>=2)break; a=a-b; swap(a,b); flag=!flag; } if(flag)printf("Stan wins\n"); else printf("Ollie wins\n"); } return 0; }