UVA-489 Hangman Judge

In “Hangman Judge,” you are to write a program that judges a series of Hangman games. For each game, the answer to the puzzle is given as well as the guesses. Rules are the same as the classic game of hangman, and are given as follows:

1. The contestant tries to solve to puzzle by guessing one letter at a time.

2. Every time a guess is correct, all the characters in the word that match the guess will be “turned over.” For example, if your guess is ‘o’ and the word is “book”, then both ‘o’s in the solution will be counted as “solved”.

3. Every time a wrong guess is made, a stroke will be added to the drawing of a hangman, which needs 7 strokes to complete. Each unique wrong guess only counts against the contestant once.

4. If the drawing of the hangman is completed before the contestant has successfully guessed all the characters of the word, the contestant loses.

5. If the contestant has guessed all the characters of the word before the drawing is complete, the contestant wins the game.

6. If the contestant does not guess enough letters to either win or lose, the contestant chickens out. Your task as the “Hangman Judge” is to determine, for each game, whether the contestant wins, loses, or fails to finish a game.

Input
Your program will be given a series of inputs regarding the status of a game. All input will be in lower case. The first line of each section will contain a number to indicate which round of the game is being played; the next line will be the solution to the puzzle; the last line is a sequence of the guesses made by the contestant. A round number of ‘-1’ would indicate the end of all games (and input).

Output
The output of your program is to indicate which round of the game the contestant is currently playing as well as the result of the game. There are three possible results: You win. You lose. You chickened out.

Sample Input
1

cheese

chese

2

cheese

abcdefg

3

cheese

abcdefgij

-1

Sample Output
Round 1

You win.

Round 2

You chickened out.

Round 3

You lose.

 

这题的难点有以下几个：

1. 理解：“You Chickened Out.”是一种什么情况；
2. 实现“猜一个已经猜过的字母算错”；
3. 维护较多的变量；

解决：

1. “You Chickened Out.”是指没有猜完就放弃了（还没死但也还没猜对就按下了回车键）；
2. “猜一个已经猜过的字母算错”可以使用将已经猜过的字母变成空格‘ ’来实现（将该语句放在循环中，一旦所猜字符串中有和原字符串相同的字母，便将原字符串中所有该字母赋值成空格， 这样再     次读入这个字母时，会找不到而判错）；
3. 使用全局变量；

代码如下：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define maxn 100
 4 int left, chance;
 5 char s[maxn], s2[maxn];
 6 int win, lose;
 7  
 8 void guess(char ch){
 9     int bad = 1;
10     for(int i=0; i<strlen(s); i++){
11         if(s[i] == ch){
12             left--;
13             s[i] = ' ';
14             bad = 0;
15         }
16     }
17     if(bad) --chance;
18     if(!chance) lose = 1;
19     if(!left) win = 1;
20 }
21  
22 int main()
23 {
24     int rnd;
25     while(scanf("%d", &rnd) && rnd != -1){
26         scanf("%s", s);
27         getchar();
28         scanf("%s", s2);
29         printf("Round %d\n", rnd);
30         win = lose = 0;
31         left = strlen(s);
32         chance = 7;
33         for(int i=0; i<strlen(s2); i++){
34             guess(s2[i]);
35             if(win || lose) break;
36         }
37         if(win) printf("You win.\n");
38         else if(lose) printf("You lose.\n");
39         else printf("You chinkened out.\n");
40     }
41     return 0;
42 }