ACM ICPC East Central North America 1994_Simply Syntax

  写这篇博客，主要是贴一段代码。该代码用于解决哈尔滨工业大学（我的母校）acm上的一道题目。闲话少叙：

  代码中含有题目要求，代码如下：

/*      Author : Li Pan      Date : 2013/2/25      Function : A program to solve an acm program named Simply Syntax. Its requirements is as follows :             0. The only characters in the language are the characters p through z and N, C, D, E, and I. 1. Every character from p through z is a correct sentence. 2. If s is a correct sentence, then so is Ns. 3. If s and t are correct sentences, then so are Cst, Dst, Est, and Ist. 4. Rules 0. to 3. are the only rules to determine the syntactical correctness of a sentence.  */#include <stdio.h>char test_string[1024];char yes_or_no;int get_last_index_of_sentence(int start) {    char current_char;    int last_index;    int temp_index;    if (start >= strlen(test_string)) {last_index = -1;return last_index;    }     current_char = test_string[start];    switch(current_char) {        case        'p' : case 'q' : case       'r' : case 's' : case 't' : case 'u' : case 'v' : case 'w' : case 'x' : case 'y' : case 'z' : last_index = start;break;case 'N' : last_index = get_last_index_of_sentence(start + 1);break;case 'C' : case 'D' : case 'E' : case 'I' :temp_index = get_last_index_of_sentence(start + 1);last_index = get_last_index_of_sentence(temp_index + 1);break;default : /*very important for forbidden characters.*/last_index = -1;break;    }    return last_index;}int main() {int i;int last_index;fscanf(stdin, "%s", test_string);last_index = get_last_index_of_sentence(0);if (-1 == last_index)fprintf(stdout, "no\n");else if (last_index < strlen(test_string) - 1)fprintf(stdout, "no\n");else fprintf(stdout, "yes\n");return 0;}