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  • KMP——Game

    Problem 2275 Game

    Accept: 41    Submit: 109
    Time Limit: 1000 mSec    Memory Limit : 262144 KB

    Problem Description

    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 
     
    #include <cstdio>
    #include <iostream>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    char s[1000005],t[1000005],tmp[1000005];
    int nextl[1000005];
    int ls,lt;
    void getnext()
    {
        nextl[0]=-1;
        for(int i=1;i<lt;i++)
        {
            int j=nextl[i-1];
            while(t[j+1]!=t[i]&&j>-1)
                j=nextl[j];
            nextl[i]=(t[j+1]==t[i])?j+1:-1;
        }
    }
    int kmp(char *a,char *b)
    {
        getnext();
        int sum=0,i=0,j=0;
        while(i<ls&&j<lt)
        {
            if(j==-1||a[i]==b[j])
                i++,j++;
            else
                j=nextl[j];
        }
        if(j==lt)
            return 1;
        return 0;
    }
    int main()
    {
        int n;
        cin>>n;
        while(n--)
        {
            cin>>s>>t;
            ls=strlen(s),lt=strlen(t);
            if(ls<lt)
            {
                printf("Bob
    ");
                continue;
            }
            if(kmp(s,t)||!strcmp(t,"0"))
            {
                printf("Alice
    ");
                continue;
            }
            reverse(t,t+lt);
            if(kmp(s,t))
            {
                printf("Alice
    ");
                continue;
            }
            printf("Bob
    ");
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/onlyli/p/7225295.html
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