ACM HDU 1517 A Multiplication Game(博弈)
A Multiplication Game
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1275 Accepted Submission(s): 773
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.
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
Recommend
LL
/*
HDU 1517
博弈题;
题意:2 个人玩游戏,从 1 开始,轮流对数进行累乘,直到超过一个指定的值。
解题思路:
如果输入是 2 ~ 9 ,因为Stan 是先手,所以Stan 必胜
如果输入是 10~18 ,因为Ollie 是后手,不管第一次Stan 乘的是什么,Stan肯定在 2 ~ 9 之间,
如果Stan乘以 2 ,那么Ollie就乘以 9 ,就到18了,如果Stan乘以 9 ,
那么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
............
必胜态是对称的!!!
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,如果1<M<=9则先手胜利,即Stan wins.如果9<M<=18
则后手胜利.
*/
#include<stdio.h>
int main()
{
double n;
while(scanf("%lf",&n)!=EOF)
{
while(n>18)n/=18;
if(n<=9) printf("Stan wins.\n");
else printf("Ollie wins.\n");
}
return 0;
}
人一我百!人十我万!永不放弃~~~怀着自信的心,去追逐梦想