hdu-1517-博弈

A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6917    Accepted Submission(s): 3930

Problem Description

Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p = 1, does his multiplication, then Ollie multiplies the number, then Stan and so on. Before a game starts, they draw an integer 1 < n < 4294967295 and the winner is who first reaches p >= n.

 

Input

Each line of input contains one integer number n.

 

Output

For each line of input output one line either 

Stan wins. 

or 

Ollie wins.

assuming that both of them play perfectly.

 

Sample Input

162 17 34012226

 

Sample Output

Stan wins. Ollie wins. Stan wins.

 

Source

University of Waterloo Local Contest 2001.09.22

 

　　　　 我们可以逆推出必胜区间和必败区间，一段一段递推，当边界到达1的时候输出此时的状态就好了，初始化L=n,op=LOSE,

当上一个区间状态是LOSE的时候，就执行 L=ceil(L/9) op=WIN

反之执行  L=ceil(L/2) op=LOSE.

 

只要记住1只要有一个子状态到0，而0的所有子状态都必须是1即可。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     long long n;
 5     while(cin>>n){
 6         bool op=0;
 7         while(n>1){
 8             if(op==0){
 9                 op=1;
10                 n=ceil((double)n/9);
11             }
12             else{
13                 op=0;
14                 n=ceil((double)n/2);
15             }
16             
17         }
18         op?puts("Stan wins."):puts("Ollie wins.");
19     }
20     return 0;
21 }