poj 2505 A multiplication game

最新推荐文章于 2021-09-10 16:36:09 发布

禾烟雨

最新推荐文章于 2021-09-10 16:36:09 发布

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文章标签：博弈

本文链接：https://blog.youkuaiyun.com/mandiheyanyu/article/details/82889340

本文深入探讨了一款基于乘法的博弈游戏算法，通过交替使用2到9之间的数字进行乘法运算，目标是在达到预设阈值前赢得比赛。文章详细分析了游戏策略，提供了实现代码，并展示了如何通过优化策略来确定胜者。

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poj 2505 A multiplication game

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.

Ollie wins.

assuming that both of them play perfectly.
Sample Input
162
17
34012226
Sample Output
Stan wins.
Ollie wins.
Stan wins.

其实这个题蛮像那种一个圆盘那棋子的，总那对面的一个，不过这里是轮流*9和2。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;

int main(){
	int n,flag;
	double sum;
	while(scanf("%d",&n)!=EOF){
		sum=1;
		flag=1;
		while(sum<n){
			if(flag){
				sum*=9;
				flag=0;
			}
			else{
				sum*=2;
				flag=1;
			}
		}
		if(flag==0){
			printf("Stan wins.\n");
		}
		else{
			printf("Ollie wins.\n");
		}
	}
	return 0;
}