FZU - 2275 (KMP)

最新推荐文章于 2019-05-08 23:29:54 发布

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最新推荐文章于 2019-05-08 23:29:54 发布

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分类专栏： KMP 文章标签： KMP

本文链接：https://blog.youkuaiyun.com/a1046765624/article/details/80112509

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本文介绍了一个简单的游戏博弈问题，通过KMP算法实现游戏胜负的快速判断。游戏中两名玩家Alice和Bob轮流操作自己的数字，目标是使两个数字相等。文章详细展示了如何使用KMP算法进行比对，并给出了完整的代码实现。

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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

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.

1A的一道KMP。虽然很简单，但也激动死我了。。哈哈哈。

思路：

主要是有一个特判，当B串，只有一个，并且是0的时候，一定是Alice 赢。其他就是利用KMP，来进行判断输出了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char a[110000],b[110000],c[110000];
int cnt;
int f[110000];
void find(char *T,char *P,int *f)
{
    int n=strlen(T);
    int m=strlen(P);
    int j=0;
    for(int i=0;i<n;i++)
    {
        if(j && T[i]!=P[j]) j=f[j];
        if(T[i]==P[j]) j++;
        if(j==m) cnt++;
    }
}
void getFail(char *P,int *f)
{
    int m =strlen(P);
    f[0]=f[1]=0;
    for(int i=1;i<m;i++)
    {
        int j=f[i];
        while(j && P[i]!=P[j]) j=f[j];
        f[i+1] = P[i]==P[j]?j+1:0;
    }
}
int work(char *q,char *w,int *e)
{
    cnt=0;
    getFail(w,e);
    find(q,w,e);
    return cnt;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        cnt=0;
        memset(f,0,sizeof(f));
        memset(c,0,sizeof(c));
        scanf("%s%s",a,b);
        int la=strlen(a);
        int lb=strlen(b);
        if(lb==1&&b[0]=='0')printf("Alice\n");
        else
        {
        if(la<lb)
        {
            printf("Bob\n");
        }
        else
        {
            for(int i=0;i<lb;i++)
            {
                c[i]=b[lb-i-1];
            }
            //cout<<c<<endl;
            int ans1=work(a,b,f);
            memset(f,0,sizeof(f));
            int ans2=work(a,c,f);
            if(ans1||ans2)
            {
                printf("Alice\n");
            }
            else printf("Bob\n");
        }
        }
    }
    return 0;
}