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.

這題的難點有以下幾個：

1. 理解：“You Chickened Out.”是一種什麼情況；
2. 實現“猜一個已經猜過的字母算錯”；
3. 維護較多的變數；

解決：

1. “You Chickened Out.”是指沒有猜完就放棄了（還沒死但也還沒猜對就按下了回車鍵）；
2. “猜一個已經猜過的字母算錯”可以使用將已經猜過的字母變成空格‘ ’來實現（將該語句放在迴圈中，一旦所猜字串中有和原字串相同的字母，便將原字串中所有該字母賦值成空格， 這樣再     次讀入這個字母時，會找不到而判錯）；
3. 使用全域性變數；

程式碼如下：

 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 }