hdu 1517 A Multiplication Game 段sg 博弈难度:0-优快云博客

本文探讨了乘法游戏的胜利策略，并提供了一个高效算法实现。通过分析数值区间内的胜负规律，玩家可以准确预测谁将赢得游戏。利用数学原理简化输入到输出的过程，实现了一种快速判断胜负的方法。

A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3832 Accepted Submission(s): 2183

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.

Source

University of Waterloo Local Contest 2001.09.22

打了sg表...1 肯定必败 (2,9)必胜 (10,18)必败,(19,36)必胜...总之就是((2*9)^i+1,(2*9)^i*9)必败,其他必胜,总结规律之后,只要判断在哪个区间即可

数据范围本来以为会爆longlong,但是*9没有爆longlong

#include <cstdio>
#include <cstring>
using namespace std;
long long n;
int main(){
    while(scanf("%I64d",&n)!=EOF){
        long long tn=1;
        bool fl=false;
        while(n>tn){
            tn*=9;fl=true;
            if(n<=tn)break;
            tn*=2;fl=false;
        }
        if(fl)printf("Stan wins.\n");
        else printf("Ollie wins.\n");
    }
    return  0;
}

转载于:https://www.cnblogs.com/xuesu/p/4104185.html