Euclid's Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2074 Accepted Submission(s): 924
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
题目的意思：
有两个正整数数，a,b 轮流减去两个数min(a,b)中的倍数，谁最后得到o谁就won,注意两个人都可以对这两个数具有同样的操作......
代码：
开始写一个决策树，不带路径压缩，然后玛德，无限tle...
超时的代码：
1 #include<cstring> 2 #include<cstdio> 3 bool flag=false; 4 int cont=1; 5 void botree(int n,int m){ 6 if(n<m)n^=m^=n^=m; 7 if(n%m) 8 for(int i=1;i*m<n;i++){ 9 cont++; 10 botree(n-i*m,m); 11 } 12 else { 13 if(cont&1) flag=true; 14 cont=1; 15 } 16 } 17 int main(){ 18 int n,m; 19 //freopen("test.in","r",stdin); 20 while(scanf("%d%d",&n,&m),n+m){ 21 flag=false; 22 botree(n,m); 23 if(flag) printf("Stan wins "); 24 else printf("Ollie wins "); 25 } 26 return 0; 27 }
然后进行了优化之后.....得到这样的代码：
1 /*Problem : 1525 ( Euclid's Game ) Judge Status : Accepted 2 RunId : 11528629 Language : C++ Author : huifeidmeng 3 Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta*/ 4 5 #include<cstring> 6 #include<cstdio> 7 bool flag=false; 8 int cont=1,cc=0; 9 void botree(int n,int m){ 10 if(n<m) n^=m^=n^=m; 11 if(m>0&&((n%m)&&n/m==1)){ 12 cont++; 13 botree(n%m,m); 14 } 15 else if(cont&1) flag=true; 16 } 17 int main(){ 18 int n,m; 19 //freopen("test.in","r",stdin); 20 while(scanf("%d%d",&n,&m),n+m){ 21 flag=false; 22 cont=1; 23 botree(n,m); 24 if(flag) printf("Stan wins "); 25 else printf("Ollie wins "); 26 } 27 return 0; 28 }