zoukankan      html  css  js  c++  java
  • Game

    Alice and Bob is playing a game.

    Each of them has a number. Alice’s number is A, and Bob’s number is B.

    Each turn, one player can do one of the following actions on his own number:

    1. Flip: Flip the number. Suppose X = 123456 and after flip, X = 654321

    2. Divide. X = X/10. Attention all the numbers are integer. For example X=123456 , after this action X become 12345(but not 12345.6). 0/0=0.

    Alice and Bob moves in turn, Alice moves first. Alice can only modify A, Bob can only modify B. If A=B after any player’s action, then Alice win. Otherwise the game keep going on!

    Alice wants to win the game, but Bob will try his best to stop Alice.

    Suppose Alice and Bob are clever enough, now Alice wants to know whether she can win the game in limited step or the game will never end.

    Input

    First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.

    For each test case: Two number A and B. 0<=A,B<=10^100000.

    Output

    For each test case, if Alice can win the game, output “Alice”. Otherwise output “Bob”.

    Sample Input

    4
    11111 1
    1 11111
    12345 54321
    123 123
    

    Sample Output

    Alice

    Bob
    Alice
    Alice
    题解:此题分为3种情况。
    (1):如果strlen(a)<strlen(b),Bob win;
    (2):如果b=='0',Alice win;
    (3) :如果b是a的子串||reverse(b)是a的子串 Alice win,反之 Bob win;(kmp 裸题);
     1 #include <iostream>
     2 #include <stdio.h>
     3 #include <string.h>
     4 #include <algorithm>
     5 using namespace std;
     6 int m1,d;
     7 char a[1000000],b[1000000];
     8 int f[1000001];
     9 void getnt()
    10 {
    11     memset(f,'0',sizeof(f));
    12     int i=0,j=-1;
    13     f[0]=-1;
    14     while(i<d)
    15     {
    16         if(j==-1||b[i]==b[j])
    17         {
    18             i++;j++;
    19             if(b[i]!=b[j])
    20                 f[i]=j;
    21             else
    22                 f[i]=f[j];
    23         }
    24         else
    25             j=f[j];
    26 
    27     }
    28 }
    29 int kmp()
    30 {
    31     getnt();
    32     int i=0, j=0;
    33     while(i<d)
    34     {
    35         if(j==-1||a[i] ==b[j])
    36             i++, j++;
    37         else
    38             j=f[j];
    39         if(j==m1)
    40             return i-m1+1;
    41     }
    42     return -1;
    43 }
    44 int main()
    45 {
    46     int T,flog;
    47     scanf("%d",&T);
    48     for(int i=0;i<T;i++)
    49     {
    50         scanf(" %s %s",&a,&b);
    51         if(b[0]=='0')
    52             printf("Alice
    ");
    53         else
    54         {
    55         if(strlen(a)<strlen(b))
    56             printf("Bob
    ");
    57         else
    58         {
    59             char p;
    60             m1=strlen(b);
    61             d=strlen(a);
    62             flog=kmp();
    63             if(flog==-1)
    64             {
    65                 for(int j=0;j<m1/2;j++)
    66                 {
    67               swap(b[j],b[m1-j-1]);                      
    68               }
    69                 flog=kmp();
    70             }
    71             if(flog==-1)
    72                 printf("Bob
    ");
    73             else
    74                 printf("Alice
    ");
    75         }
    76         }
    77     }
    78     return 0;
    79 }

    Hint

    For the third sample, Alice flip his number and win the game.

    For the last sample, A=B, so Alice win the game immediately even nobody take a move.

  • 相关阅读:
    CentOS7和Ubuntu18.10下运行Qt Creator出现cannot find -lGL的问题的解决方案
    C++中使用CMake编译管理项目
    C++ std::isnan等函数的使用
    七个开源的 Spring Boot 前后端分离项目,一定要收藏!
    基于Select模型的Windows TCP服务端和客户端程序示例
    简单的C++11线程池实现
    Windows下FFMEPG编译
    MIMO431学习笔记目录
    golang调用海康sdk
    GRPC-go版本
  • 原文地址:https://www.cnblogs.com/moomcake/p/8975856.html
Copyright © 2011-2022 走看看