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poj 2505 A multiplication game(博弈)

时间:2016-01-30 22:16:12      阅读:242      评论:0      收藏:0      [点我收藏+]

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A multiplication game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 5622   Accepted: 2811

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

 

【思路】

       博弈

       2<=n<=9,先手必胜

       10<=n<=18,先手必败

       19<=n<=162,先手必胜

       163<=n<=324,先手必败

       …

       =-=

【代码】

 

 1 #include<cstdio>
 2 
 3 int main() {
 4     double n;
 5     while(scanf("%lf",&n)==1) {
 6         while(n>18) n/=18;
 7         if(n<=9) puts("Stan wins.");
 8         else puts("Ollie wins.");
 9     }
10     return 0;
11 }

 

poj 2505 A multiplication game(博弈)

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原文地址:http://www.cnblogs.com/lidaxin/p/5171496.html

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