Usaco 1.3.3 calf FLAC (calfflac)

1. Silly brute-force search: The two pointers start from the first and last pointers and move them in the middle to determine whether the string between the two pointers is the number of input packets. Too many repeated searches cause the last test data to time out:

/* <Br/> ID: gengjia1 <br/> LANG: C <br/> task: calfflac <br/> */<br/> # include <stdio. h> <br/> # include <stdlib. h> <br/> # include <string. h> <br/> # include <ctype. h> <br/> // # define ndebug <br/> # include <assert. h> </P> <p> # define max_len 20000 <br/> int count; </P> <p> struct STR {<br/> int Len; <br/> char s [max_len + 1]; <br/>}; </P> <p> struct pal {<br/> int Len; <br/> int begin; <br/> int end; <br/>}; </P> <p> int lett Er (const char a) <br/>{< br/> If (A> = 'A' & A <= 'Z ') | <br/> (A> = 'A' & A <= 'Z') <br/> return 1; <br/> return 0; <br/>}</P> <p> int main (void) {<br/> file * fin = fopen ("calfflac. in "," R "); <br/> file * fout = fopen (" calfflac. out "," W "); <br/> assert (Fin! = NULL & fout! = NULL); </P> <p> struct str s = {0}; <br/> struct pal Pa = {0 }; <br/> char TMP [80] = {0}; <br/> char T [max_len] = {0}; <br/> int curlen = 0; <br/> int I, j, k, p, q; </P> <p> while (fgets (TMP, 80, fin )! = NULL) <br/>{< br/> strcat (S. s, TMP); <br/>}< br/> S. len = strlen (S. s); </P> <p> for (I = 0; I <S. len; I ++) <br/>{< br/> If (letter (S. s [I]) = 0) <br/> continue; <br/> for (j = S. len-1; j> I; j --) <br/>{< br/> If (letter (S. s [J]) = 0) <br/> continue; <br/> curlen = 0; <br/> for (P = I, q = J; P <q;) <br/>{< br/> If (letter (S. s [p]) = 0) <br/>{< br/> P ++; <br/> continue; <br/>}< br/> If (letter (S. s [Q]) = 0) <br/>{< br/> q --; <br/> continue; <br/>}< br/> If (tolower (S. s [p]) = tolower (S. s [Q]) <br/>{< br/> curlen + = 2; <br/> P ++; <br/> q --; <br/>}< br/> else <br/> {<br/> curlen = 0; <br/> break; <br/>}< br/> If (P = Q & letter (S. s [p]) = 1) <br/> curlen + = 1; </P> <p> If (curlen> Pa. len) <br/> {<br/> Pa. len = curlen; <br/> Pa. begin = I; <br/> Pa. end = J; <br/>}< br/> fprintf (fout, "% d/N", Pa. len); <br/> for (I = pa. begin; I <= pa. end; I ++) <br/>{< br/> fprintf (fout, "% C", S. s [I]); <br/>}< br/> fprintf (fout, "/N"); </P> <p> fclose (FIN ); <br/> fclose (fout); <br/> exit (0); <br/>}< br/>