zoukankan      html  css  js  c++  java
  • LightOj 1020 博弈

    思路介绍:

           1. 如果首先由Alice取,定义ans[i],如果ans[i]=1表示Alice会取胜,反之Bob取胜。枚举前100项,ans[1]=0,ans[2]=1,ans[i]=!(ans[i-1]&&ans[i-2]);

              可以发现规律:当i为2,3,5,6,8,9....时Alice取胜,所以Alice取胜的条件为:i%3!=1;

          2.如果Bob先取,ans[1]=1,ans[2]=1,ans[i]=!(ans[i-1]&&ans[i-2])

            规律:当i为1,2,4,5,7,8...时Bob取胜,、;所以Bob取胜的条件为:i%3!=0;

    1020 - A Childhood Game
    PDF (English) Statistics Forum
    Time Limit: 0.5 second(s) Memory Limit: 32 MB
    Alice and Bob are playing a game with marbles; you may have played this game in childhood. The game is playing by alternating turns. In each turn a player can take exactly one or two marbles.

    Both Alice and Bob know the number of marbles initially. Now the game can be started by any one. But the winning condition depends on the player who starts it. If Alice starts first, then the player who takes the last marble looses the game. If Bob starts first, then the player who takes the last marble wins the game.

    Now you are given the initial number of marbles and the name of the player who starts first. Then you have to find the winner of the game if both of them play optimally.

    Input
    Input starts with an integer T (≤ 10000), denoting the number of test cases.

    Each case contains an integer n (1 ≤ n < 231) and the name of the player who starts first.

    Output
    For each case, print the case number and the name of the winning player.

    Sample Input
    Output for Sample Input
    3
    1 Alice
    2 Alice
    3 Bob
    Case 1: Bob
    Case 2: Alice
    Case 3: Alice


    PROBLEM SETTER: JANE ALAM JAN

    <span style="color:#6600cc;">/**************************************
         author   : Grant Yuan
         time     : 2014/8/20 13:38
         algorithm: 博弈
         source   : LightOj 1020
    ***************************************/
    #include<bits/stdc++.h>
    #define Alice "Alice"
    #define Bob "Bob"
    
    using namespace std;
    char s[6];
    int n,t;
    int main()
    {
        scanf("%d",&t);
        for(int i=1;i<=t;i++)
        {
            memset(s,0,sizeof(s));
            scanf("%d%s",&n,s);
            printf("Case %d: ",i);
            if(strcmp(s,Alice)==0){
                if(n%3==1){
                    printf("Bob
    ");
                }
                else{
                    printf("Alice
    ");
                }
            }
            else{
                if(n%3==0){
                    printf("Alice
    ");
                }
                else{
                    printf("Bob
    ");
                }
            }
        }
        /*ans[1]=false;ans[2]=true;
        for(int i=3;i<=100;i++)
        {
            ans[i]=true;
            if(ans[i-1]&&ans[i-2])
                ans[i]=false;
        }
        for(int i=1;i<=100;i++)
        {
            printf("%d ",i);
            if(ans[i])
                printf("1
    ");
            else
                printf("0
    ");
        }*/
        return 0;
    }
    </span>


     

  • 相关阅读:
    sql server 2008数据复制方法
    排错技能:任务管理器中追踪某w3wp.exe是哪个IIS站点的application pool
    SplendidCRM中给来自EditView中的listbox控件设置选中值或数据源
    jQuery String Functions
    [转]jquery getJSON 数据联动(采用序列化和反序列化获取数据) .
    [转]javascript eval函数解析json数据时为什加上圆括号eval("("+data+")")
    深入理解C语言
    Qt回忆录之配置开发环境
    360电话面试
    浅谈C++设计模式之单例模式
  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254450.html
Copyright © 2011-2022 走看看