joj1474 Soundex Indexing

 1474: Soundex Indexing

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	246	119	Standard

The Soundex Index System was developed so that similar sounding names, or names with similar spelling could be encoded for easy retrieval. It has been used by the U.S. Bureau of the Census, and some States use it to help encode your driver's license number. Your task is to read a sequence of names, one at a time and one per line, compute the corresponding soundex code, and write to the output file the name and its soundex code (one line of output per name).

 

Names will contain from 1 to 20 upper case, alphabetic characters (ASCII values 65 thru 90, inclusive). Names shorter than 20 characters will NOT be padded with blanks. Thus a name will consist of upper case letters only.

How to generate the Soundex Code:

A Soundex Code always consists of a letter followed by three digits. Here are the rules for soundex encoding:

 

1.

The first letter of a name appears as the first and only letter in the soundex code.

2.

The letters A, E, I, O, U, Y, W and H are never encoded, but do break successive code sequences (see next rule).

3.

All other letters are encoded EXCEPT when they immediately follow a letter (including the first letter) that would be encoded with the same code digit.

4.

The soundex code guide is:

 

Code	Key Letters and Equivalents
1	B, P, F, V
2	C, S, K, G, J, Q, X, Z
3	D, T
4	L
5	M, N
6	R

 

5.

Trailing zeros are appended to short codes so all names are encoded with a letter followed by three digits.

6.

Longer codes are truncated after the third digit.

Input

The input file contains a list of names, one per line. Each name will not exceed 20 characters, and you may assume that only upper case letters will be used. Your program should continue to read names until the end of the file is detected.

Output

The output written to the file should consist of a column of names and a column of their corresponding soundex codes. Write the headings ``NAME" and ``SOUNDEX CODE" in the first line of the output file in columns 10 and 35, respectively. After the heading line, the names and soundex codes should be written (one pair per line) with the name starting in column 10 and the soundex code beginning in column 35. The comment ``END OF OUTPUT" should appear at the end of the output file on the line immediately after the last name. This comment should be written starting in column 20.

Sample Input

 

LEE

KUHNE

EBELL

EBELSON

SCHAEFER

SCHAAK

Sample Output

         NAME                     SOUNDEX CODE

         LEE                      L000

         KUHNE                    K500

         EBELL                    E140

         EBELSON                  E142

         SCHAEFER                 S160

         SCHAAK                   S200

                   END OF OUTPUT

         |         |              |

         |         |              |__ Column 35

         |         |__ Column 20

         |__ Column 10

 

 

这题真是欲仙欲死题中的典范，三次PE伤不起........

那个Column10是说那一列是第十列，也就是说前面是9个空格，9个空格啊有木有，不是10个.......

还有，写得太复杂了，能大量简化....

 

  1 #include <stdio.h>

  2 #include <string.h>

  3 

  4 int main()

  5 {

  6     //freopen("in.txt", "r", stdin);

  7     //freopen("out.txt", "w", stdout);

  8 

  9     char s[5000][22];

 10     char st[22];

 11 

 12     int n = 0;

 13     int i, j;

 14     char former, now;

 15 

 16     while (scanf("%s", &s[n++]) == 1)

 17     {

 18 

 19     }

 20 

 21     const char *name = "NAME";

 22     const char *code = "SOUNDEX CODE";

 23 

 24 

 25     for(i=0; i<9; ++i)

 26     {

 27         printf(" ");

 28     }

 29     printf("%s", name);

 30     for(int i=14; i<35; ++i)

 31     {

 32         printf(" ");

 33     }

 34     printf("%s\n", code);

 35 

 36     //printf("N=%d\n", n);

 37     for (i=0; i<n-1; ++i)

 38     {

 39 

 40         //printf("HH");

 41         int p = 0;

 42         for (j=0; j<strlen(s[i]); ++j)

 43         {

 44             //printf("strlen=%d\n", strlen(s[i]));

 45             if (0 == j)

 46             {

 47                 st[p++] = s[i][0];

 48             }

 49             if ('A'==s[i][j] || 'E'==s[i][j] || 'I'==s[i][j] || 'O'==s[i][j] || 'U'==s[i][j] || 'Y'==s[i][j] || 'W'==s[i][j] || 'H'==s[i][j])

 50             {

 51                 if (0 != j)

 52                 {

 53                     former = 'x'; //处理隔断。

 54                 }

 55                 continue;

 56             }

 57 

 58             if ('B'==s[i][j] || 'P'==s[i][j] || 'F'==s[i][j] || 'V'==s[i][j])

 59             {

 60                 if (0 == j)

 61                 {

 62                     former = '1';

 63                     continue;

 64                 }

 65                 else

 66                 {

 67                     //printf("HHHH\n");

 68                     now = '1';

 69                     //printf("for=%c\n", former);

 70                 }

 71                 st[p++] = '1';

 72             }

 73             if ('C'==s[i][j] || 'S'==s[i][j] || 'K'==s[i][j] || 'G'==s[i][j] || 'J'==s[i][j] || 'Q'==s[i][j] || 'X'==s[i][j] || 'Z'==s[i][j])

 74             {

 75                 if (0 == j)

 76                 {

 77                     former = '2';

 78                     continue;

 79                 }

 80                 else

 81                 {

 82                     now = '2';

 83                 }

 84                 st[p++] = '2';

 85             }

 86             if ('D'==s[i][j] || 'T'==s[i][j])

 87             {

 88                 if (0 == j)

 89                 {

 90                     former = '3';

 91                     continue;

 92                 }

 93                 else

 94                 {

 95                     now = '3';

 96                 }

 97                 st[p++] = '3';

 98             }

 99             if ('L'==s[i][j])

100             {

101                 if (0 == j)

102                 {

103                     former = '4';

104                     continue;

105                 }

106                 else

107                 {

108                     now = '4';

109                 }

110                 st[p++] = '4';

111             }

112             if ('M'==s[i][j] || 'N'==s[i][j])

113             {

114                 if (0 == j)

115                 {

116                     former = '5';

117                     continue;

118                 }

119                 else

120                 {

121                     now = '5';

122                 }

123                 st[p++] = '5';

124             }

125             if ('R'==s[i][j])

126             {

127                 if (0 == j)

128                 {

129                     former = '6';

130                     continue;

131                 }

132                 else

133                 {

134                     now = '6';

135                 }

136                 st[p++] = '6';

137             }

138 

139             if (now == former && j != 0)

140             {

141                 --p;

142             }

143             else

144             {

145                 former = now;

146 

147             }

148         }

149         //printf("P=%d\n", p);

150         if (p < 4)

151         {

152             for ( ; p<4; ++p)

153             {

154                 st[p] = '0';

155             }

156         }

157 

158         //输出

159 

160 

161         for(int t=0; t<9; ++t)

162         {

163             printf(" ");

164         }

165         printf("%s", s[i]);

166         for(int t=10+strlen(s[i]); t<=34; ++t)

167         {

168             printf(" ");

169         }

170         for (int k=0; k<4; ++k)

171         {

172             printf("%c", st[k]);

173         }

174         printf("\n");

175 

176     }

177 

178     for(i=0; i<19; ++i)

179     {

180         printf(" ");

181     }

182     printf("END OF OUTPUT\n");

183 

184 

185     return 0;

186 }