489 - Hangman Judge

题意:
1. 猜单词游戏, 输入一个 puzzle 和 guess, 判断 guess 是否在限定次数(7次)内猜中 puzzle 中的所在字符.
2. guess 中每一个猜错的字符, 只计算一次; 即若猜了 a, 不中, 则下次再猜 a , 仍不中, 但不会递增 stroke.
3. 若错误次数超过限定次数, 则输出 lose, 若在限定次数内猜完 puzzle, 则输出 win. 若 guess 中的字符已遍历完, 既未猜完 puzzle, 也未到达限定次数, 则输出 chickened out.

思路:
取 guess 中每个字符, 判断此字符是否在 puzzle 中, 若在, 则把 puzzle 中相同字符全部删掉; 若不在, 则判断此字符是否以前猜过, 若未猜过, 则递增 stroke, 并把错误字符添加到 wrongAnswer 中; 若已猜过, 则不作处理.

要点:
1.  C++ 字符串中查找指定字符, 使用以下代码

wrongAnswer.find(guesses[i]) == string::npos

2. C++ 字符串删除指定字符, 使用以下代码

string::iterator new_end = remove_if(puzzle.begin(), puzzle.end(), 
                                     bind2nd(equal_to<char>(), guesses[i]));
puzzle.erase(new_end, puzzle.end());

3. STL 容器(包含 string) 都可以使用 empty() 函数来判断其是否为空. 

题目:
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=94&page=show_problem&problem=430

代码:

# include <iostream>
# include <string>
# include <cstdio>
# include <cstdlib>
# include <algorithm>
# include <functional>
using namespace std;

// 以下这两种写法是有区别的, cin 对于回车换行的处理太诡异了
// 所以保险时可用 getline 读进来一个字符串, 再转成 int, 至少不会出错
// cin >> round >> blank;     
// cin >> round;
// cin >> blank;
int main(int argc, char const *argv[])
{
  #ifndef ONLINE_JUDGE
    freopen ("489_i.txt", "r", stdin);  
    freopen ("489_o.txt", "w", stdout); 
  #endif
  const int MAX_STROKES = 7;

  int round;
  string puzzle;
  string guesses;

  while (cin >> round >> puzzle >> guesses && round != -1) {

    string wrongAnswer = "";
    int strokes = 0;

    cout << "Round " << round << endl;

    for (int i=0; i<guesses.length(); i++) {
      if (puzzle.find(guesses[i]) == string::npos) {
        // 没有猜中, 则每个猜错的只计算一次 stroke
        if (wrongAnswer.find(guesses[i]) == string::npos) {
          ++strokes;
          if (strokes == MAX_STROKES) {
            cout << "You lose." << endl;
            break;
          }

          wrongAnswer.push_back(guesses[i]);
        }
      } else {
        // 在 puzzle 中删除猜中了的 guesses[i]
        string::iterator new_end = remove_if(puzzle.begin(), puzzle.end(), 
                                             bind2nd(equal_to<char>(), 
                                                     guesses[i]));
        puzzle.erase(new_end, puzzle.end());

        if (puzzle.empty()) {
          cout << "You win." << endl;
          break;
        }
      }
    }

    if (!puzzle.empty() && strokes != MAX_STROKES) {
      cout << "You chickened out." << endl;
    }
  }
  return 0;
}

环境: C++ 4.5.3 - GNU C++ Compiler with options: -lm -lcrypt -O2 -pipe -DONLINE_JUDGE