题意:2 个人玩游戏,给定一个数n,从 1 开始,轮流对数进行累乘一个数(2~9中取),
直到第一次等于或超过n为赢.
思路:1)找规律
如果n是 2 ~ 9 ,Stan 必胜。
如果输入是 10~18 ,不管第一次Stan 乘的是什么,Stan肯定在 2 ~ 9 之间,
无论stan乘以什么,Ollie乘以大于1的数都都能超过 10 ~ 18 中的任何一个数。Ollie 必胜。
如果输入是 19 ~ 162,那么这个范围是 Stan 的必胜态。
如果输入是 163 ~ 324 ,这是又是Ollie的必胜态。
............
必胜态是有规律可循的。
如果"我方"首先给出了一个在n不断除18后的得到不足18的
数m,"我方"就可以取得胜利,然而双方都很聪明,所以这样胜负就决定于n了,
如果n不断除18后的得到不足18的数m,
若1<m<=9则先手胜利,
若9<m<=18则后手胜利.
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.
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.
一道巴什博弈的转化题目其实差不多以前是加法现在是乘法了,不过没有想到用double导致错了好多次。。。。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
double n;
while(~scanf("%lf",&n))
{
while(n>18)
{
n/=18;
}
if(n<=9)
printf("Stan wins.\n");
else
printf("Ollie wins.\n");
}
return 0;
}