AtCoder Beginner Contest 045 B - 3人でカードゲームイージー / Card Game for Three (ABC Edit)...

本文介绍了一个简单的三人纸牌游戏算法实现,通过模拟游戏过程来判断最终赢家。玩家依次弃掉顶部带有特定字母的纸牌,根据字母决定下一位玩家,直至某玩家牌堆为空即为胜者。

Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

Alice, Bob and Charlie are playing Card Game for Three, as below:

  • At first, each of the three players has a deck consisting of some number of cards. Each card has a letter ab or c written on it. The orders of the cards in the decks cannot be rearranged.
  • The players take turns. Alice goes first.
  • If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says a, Alice takes the next turn.)
  • If the current player's deck is empty, the game ends and the current player wins the game.

You are given the initial decks of the players. More specifically, you are given three strings SASB and SC. The i-th (1≦i≦|SA|) letter in SA is the letter on the i-th card in Alice's initial deck. SB and SC describes Bob's and Charlie's initial decks in the same way.

Determine the winner of the game.

Constraints

  • 1≦|SA|≦100
  • 1≦|SB|≦100
  • 1≦|SC|≦100
  • Each letter in SASBSC is ab or c.

Input

The input is given from Standard Input in the following format:

SA
SB
SC

Output

If Alice will win, print A. If Bob will win, print B. If Charlie will win, print C.


Sample Input 1

Copy
aca
accc
ca

Sample Output 1

Copy
A

The game will progress as below:

  • Alice discards the top card in her deck, a. Alice takes the next turn.
  • Alice discards the top card in her deck, c. Charlie takes the next turn.
  • Charlie discards the top card in his deck, c. Charlie takes the next turn.
  • Charlie discards the top card in his deck, a. Alice takes the next turn.
  • Alice discards the top card in her deck, a. Alice takes the next turn.
  • Alice's deck is empty. The game ends and Alice wins the game.

Sample Input 2

Copy
abcb
aacb
bccc

Sample Output 2

Copy
C

 题解:不知道为啥队列就会哇  还是直接模拟

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#include <queue>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll gcd(ll a,ll b){
    return b?gcd(b,a%b):a;
}
bool cmp(int x,int y)
{
    return x>y;
}
const int N=55;
const int mod=1e9+7;
char A[101],B[101],C[101];
int main()
{
    int a,b,c;
    int la,lb,lc;
    while(cin>>A>>B>>C){
        int tmp;
        la=strlen(A),lb=strlen(B),lc=strlen(C);
        tmp=1,a=b=c=-1;
        while(1){
            if(a==la){
                cout<<"A"<<endl;
                break;
            }
            else if(b==lb){
                cout<<"B"<<endl;
                break;
            }
            else if(c==lc){
                cout<<"C"<<endl;
                break;
            }
            if(tmp==1){
                a++;
                tmp=A[a]-'a'+1;
            }
            else if(tmp==2){
                b++;
                tmp=B[b]-'a'+1;

            }
            else if(tmp==3){
                c++;
                tmp=C[c]-'a'+1;
            }
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/shixinzei/p/7289545.html

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