fzu Problem 2275 Game（kmp）

最新推荐文章于 2019-05-08 23:29:54 发布

转载最新推荐文章于 2019-05-08 23:29:54 发布 · 149 阅读

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原文链接：http://www.cnblogs.com/stepping/p/7227822.html

本文介绍了一个简单的游戏博弈问题，通过KMP算法判断玩家Alice是否能在有限步骤内赢得游戏。问题中涉及两个玩家轮流操作自己的数字，目标是使两数字相等。文章提供了完整的代码实现，并解释了关键的算法思路。

Problem 2275 Game

Accept: 62 Submit: 165
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

11111 1

1 11111

12345 54321

123 123

Sample Output

Alice

Bob

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.

Source

第八届福建省大学生程序设计竞赛-重现赛（感谢承办方厦门理工学院）

题解：kmp

如果b的位数大于a，B获胜。如果B为0，A获胜。如果b是a的逆序，A获胜，其他情况，如果b是a或者逆序a的子串，A获胜。

#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 T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%s",&s,&t);
        ls = strlen(s);
        lt = strlen(t);
        if (lt>ls) { printf("Bob\n"); continue; }
        if(kmp(s,t) || !strcmp(t,"0")) {printf("Alice\n"); continue;}  //如果t字符串是0，那么一定是Alice赢
        reverse(t,t+lt);  //这个翻转用法第一次碰到
        if(kmp(s,t)) {printf("Alice\n"); continue;}
        printf("Bob\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/stepping/p/7227822.html