A Multiplication Game HDOJ (巴什)

最新推荐文章于 2018-12-16 12:41:00 发布

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Stan与Ollie进行了一场独特的数字乘法比赛，通过不断将初始数值1乘以2到9之间的任意整数，以达到或超过预先设定的目标数字n来决出胜负。本文探讨了如何利用简单的数学逻辑和循环来解决这个问题，并提供了高效求解策略。

A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4369 Accepted Submission(s): 2480

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.

这种相乘的类似于相加的,但都是巴什，都是由简单的问题推想复杂的

之前的选手可以拿1~m个石子，当先手拿得石子m满足n%(m+1)==0时后手必赢。

相同的此题的分析方法也类似：

当n->2~9时先手赢

n->10~时后手赢。以此类推

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
  double  m,n;
    ios::sync_with_stdio(false);
    while(cin>>n)
    {
        if(n<=9)
        {
            puts("Stan wins.");
        }
        else
        {
            while( n>18)//跃过多次循环问题简化
            {
                n=n/18;
            }
            if(n<=9)
            puts("Stan wins.");
            else
            puts("Ollie wins.");
        }
    }
    return 0;
}