http://codeforces.com/gym/101190/attachments
Problem A.Abbreviation
Input file: abbreviation.in
Output file: abbreviation.out
An abbreviation (from Latin brevis, meaning short) is a shortened form of a word or phrase. In this problem you must write an automated tool that replaces a sequence of capitalized words with the corresponding abbreviation that consists of the first upper case letters only, followed by a full definition in parenthesis. See sample input and output.
Let us make some formal definitions. A word in a text is a maximally long sequence of lower and upper case English letters. A capitalized word is a word that consists of an upper case letter followed by one or more lower case letters. For example, “Ab”, “Abc”, “Abcd”, and “Abcde“ are all capitalized words, while “ab”, “A”, “AB“, “ABc“ and “AbC“ are not.
An abbreviatable sequence of words is a sequence of two or more capitalized words that are separated by exactly one space, no line breaks or punctuation are allowed inside it.
An abbreviation of an abbreviatable sequence of words is a sequence of the first (upper case) letters of each word, followed by a single space, an opening parenthesis, the original abbreviatable sequence, and a closing parenthesis.
Input
The input file consists of up to 1 000 lines of text with up to 120 characters on each line. Each line consists of spaces, upper and lower case letters, commas or dots. There are no leading or trailing spaces on lines and there are no empty lines. There is at least one line in the input file.
Output
Write to the output file the original text with every abbreviatable sequence of words replaced with the corresponding abbreviation.
Examples
abbreviation.in
This is ACM North Eastern European Regional Contest,
sponsored by International Business Machines.
The. Best. Contest. Ever.
A Great Opportunity for all contestants.
abbreviation.out
This is ACM NEERC (North Eastern European Regional Contest),
sponsored by IBM (International Business Machines).
The. Best. Contest. Ever.
A GO (Great Opportunity) for all contestants.
abbreviation.in
ab Ab A Abc AB Abcd ABc Abcde AbC
abbreviation.out
ab Ab A Abc AB Abcd ABc Abcde AbC
abbreviation.in
Oh No Extra Spaces.And,Punctuation Ruin Everything
abbreviation.out
Oh No ES (Extra Spaces).And,PRE (Punctuation Ruin Everything)
我的思路是把单词和单词间隔分别提出来，存到数组b和c，后期好处理一点，细节很多，需要耐心
1 #include <stdio.h> 2 #include <string.h> 3 4 int judge_xiao(char ch) 5 { 6 if(ch>='a'&&ch<='z') return 1; 7 else return 0; 8 } 9 10 int judge_da(char ch) 11 { 12 if(ch>='A'&&ch<='Z') return 1; 13 else return 0; 14 } 15 16 int len, i, curb, curc, j, k; 17 int book[10005], book1[10005]; 18 char a[10005], b[10005][10005], c[10005][10005]; 19 20 int main() 21 { 22 freopen("abbreviation.in", "r", stdin); 23 freopen("abbreviation.out", "w", stdout); 24 while(gets(a)) 25 { 26 memset(book, 0, sizeof(book)); 27 memset(book1, 0, sizeof(book1)); 28 len = strlen(a); 29 curb = 0; 30 curc = 0; 31 32 for(i=0; i<len; i++) 33 { 34 j = 0; 35 while((judge_xiao(a[i])||judge_da(a[i]))&&i<len) 36 { 37 b[curb][j++] = a[i]; 38 i++; 39 } 40 b[curb][j] = ' '; 41 curb++; 42 if(i<len) 43 { 44 j = 0; 45 while((judge_xiao(a[i])==0&&judge_da(a[i])==0)&&i<len) 46 { 47 c[curc][j++] = a[i]; 48 i++; 49 } 50 c[curc][j] = ' '; 51 curc++; 52 i--; 53 } 54 55 } 56 57 for(i=0; i<curb; i++) 58 { 59 len = strlen(b[i]); 60 if(judge_da(b[i][0])==0||len==1) continue; 61 else 62 { 63 j = 1; 64 while(judge_xiao(b[i][j])==1&&j<len) 65 { 66 j++; 67 } 68 if(j>=len) book[i] = 1; 69 } 70 } 71 72 for(i=0; i<curc; i++) 73 { 74 if(strcmp(c[i], " ")==0) book1[i] = 1; 75 } 76 77 78 book1[curc] = 1; 79 int num; 80 for(i=0; i<curb; i++) 81 { 82 k = i; 83 num = 0; 84 while(book[k]==1&&k<curb) 85 { 86 num++; 87 if(book1[k]==0) break; 88 k++; 89 } 90 if(book[k]==0||k>=curb) k--; 91 if(num>1) 92 { 93 for(j=i; j<=k; j++) 94 { 95 printf("%c", b[j][0]); 96 } 97 printf(" "); 98 printf("("); 99 for(j=i; j<=k; j++) 100 { 101 if(j==k) printf("%s)", b[j]); 102 else printf("%s", b[j]); 103 if(j<curc) printf("%s", c[j]); 104 } 105 i = k; 106 } 107 else 108 { 109 printf("%s", b[i]); 110 if(i<curc) printf("%s", c[i]); 111 } 112 } 113 printf(" "); 114 } 115 return 0; 116 }