http://acm.zcmu.edu.cn/JudgeOnline/problem.php?id=1893

1893: String Game

Time Limit: 1 Sec  

Memory Limit: 128 MB

Submit: 55  

Solved: 20

[ ][ ][ ]

Description

Alice and Bob are playing the following game with strings of letters.

Before the game begins, an initial string and a target string are decided. The initial string is at least as long as the target string. Then, Alice and Bob take turns, starting with the initial string. Bob goes first. In each turn, the current player removes either the first or the last letter of the current string. Once the length of the current string becomes equal to the length of the target string, the game stops. If the string at the end of the game is equal to the target string, Alice wins the game; otherwise Bob wins.

Determine who will win the game if both players are playing optimally.

Input

Each test case starts with 

N

, the number of inputs to process. Each input consists of one line, which contains the initial string, followed by a space, followed by the target string. Each string consists of only lowercase letters. The total input length will be less than 500000 characters.

Output

For each input, output the winner, which will either be Alice

 or Bob

.

Sample Input

5

aba b

bab b

aaab aab

xyz mnk

xyz xyz

Sample Output

Alice

Alice

Bob

Bob

Alice

大致题意是：每回合每人可以删去字符串a的第一个或者最后一个字符，直到a的长度等于b的长度，此时若a=b，则

Alice胜。Bob先走。

Alice胜的情况：

1.很明显，如果b正好在a的中间，那么无论Bob删哪边，Alice删另一侧就可以了；

2.如果在b不a的中间，Alice先和Bob删异侧的，直达遇到b，假设此时的a是（b）...，那么如果b的后面有一或二

个字符，就可以在之前的Bob删a的最后一个字符的某次和Bob删同侧的，这样最后一回合的时候就会变成x（b）y【x，y代表一个字符】，则Alice胜；但是Bob可能一直删a的第一个字符，所以同理当

Bob

只删a的第一个字符时，应出现...(b)【b的前面只有一或二个字符】。

    综上：b不a的中间时，若

删去

a的第一个字符

【

an-bn为奇数

】（或前两个字符【an-bn为

偶数】）

或者删去a的最后

一个字符【an-bn为奇数】（或最后两个字符【an-bn为

偶数】）

均能使b在a的中间，则Alice胜。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #define maxn 500055int main(){    int T;    scanf("%d",&T);    while(T--)    {        char a[maxn],b[maxn];        scanf("%s%s",a,b);        int an=strlen(a),bn=strlen(b),flag=0;        int d=(an-bn)/2;        if((an-bn)%2)        {            if(strncmp(a+d+1,b,bn)==0&&strncmp(a+d,b,bn)==0)                flag=1;        }        else        {            if(strncmp(a+d,b,bn)==0)                flag=1;            else{                if(strncmp(a+d+1,b,bn)==0&&strncmp(a+d-1,b,bn)==0)                {flag=1;}            }        }        if(flag)            printf("Alice\n");        else            printf("Bob\n");    }    return 0;}