Simple string simulation questions and simple string simulation questions

Simple string simulation questions and simple string simulation questions
Problem 2183 simple questionAccept: 48 Submit: 236
Time Limit: 1000 mSec Memory Limit: 32768 KB Problem Description

There are some simple compressed strings. For example, a [120] represents 120. For the string acb [3] d [5] e

Now we will give you two strings, cString and nString. One is a compressed string, and the other is not compressed.

Determine whether nString is a substring of cString. If it is True, otherwise False is output. the cString length clen range is 0 <clen <1000, and the nlen range of nString length is 0 <nlen <1000; cString only contains 26 lowercase letters, [], number (greater than 0 and less than 10 ^ 9 ). NString only contains 26 lower-case letters.

Sample Inputacb [3] d [5] ebd Sample OutputTrue Source

FOJ prize-winning monthly round-September

At first, I got an error and didn't know why. Later I found that the while (scanf ("% s", str1, str2) is missing )! = EOF) I feel cheated.

The idea of this question:

It is to separate the numbers and characters, and then you can judge it.

# Include <stdio. h> # include <string. h> char str1 [1500], str2 [1500]; char string1 [1234567], string2 [1234567]; // I is the starting position of the number, and j is the ending position of the number; used to calculate the number of string a; int change (int I, int j) {int ans = 0, l = 1; for (int k = j; k> = I; k --) {ans + = (str1 [k]-'0') * l; l = l * 10;} return ans;} int main () {int l1, l2; int I, j, k = 0; while (~ Scanf ("% s", str1, str2) {l1 = strlen (str1); l2 = strlen (str2); int num1 [1500] = {0 }, num2 [1500] = {0}; char c; int num = 0; for (I = 0; I <l1; I ++) {if (I = 0) {c = str1 [I]; string1 [k] = c; if (str1 [I + 1] = '[') {for (j = I + 2 ;; j ++) if (str1 [j] = ']') break; int ans = change (I + 2, J-1); num1 [k] + = ans ;} else num1 [k] ++;} else {if (str1 [I] = c) {if (str1 [I + 1] = '[') {for (j = I + 2; j ++) if (str1 [j] = ']') break; int ans = change (I + 2, J-1 ); num1 [k] + = ans;} else num1 [k] + +;} Else if (str1 [I]! = C & str1 [I]> = 'A' & str1 [I] <= 'Z') {k ++; c = str1 [I]; string1 [k] = c; if (str1 [I + 1] = '[') {for (j = I + 1; j ++) if (str1 [j] = ']') break; int ans = change (I + 2, J-1); num1 [k] = ans ;} else num1 [k] ++ ;}} string1 [k + 1] = '\ 0';/* printf ("% d \ n", len1 ); for (I = 0; I <= k; I ++) {printf ("% c % d \ n", string1 [I], num1 [I]);} */int len1 = k + 1; // change str2; (that is, string B) k = 0; for (I = 0; I <l2; I ++) {char temp = str2 [I]; int num = 1; while (str2 [I] = str2 [I + 1]) {num ++; I ++ ;} num2 [k] = Num; string2 [k] = temp; k ++;} string2 [k] = '\ 0'; int len2 = k; // represents several characters in string2; // next we will judge; // printf ("% d \ n", len1, len2); k = 0; int p, q, flag = 1; for (I = 0; I <len1; I ++) {flag = 1; if (string1 [I] = string2 [k]) {// len2 = 1; if (len2 = 1) {if (num1 [I]> = num2 [k]) {flag = 1; puts ("True"); break ;} else {flag = 0; continue;} else if (len2> 1) {for (p = k + 1, q = I + 1; p <= len2-2; p ++, q ++) {if (string1 [q]! = String2 [p]) | (string1 [q] = string2 [p] & num1 [q]! = Num2 [p]) {flag = 0; break;} // printf ("% d \ n", flag );} // printf ("% d \ n", p, q); if (flag) {if (string1 [I] = string2 [k] & num1 [I]> = num2 [k]) & (string1 [q] = string2 [p] & num1 [q]> = num2 [p]) {flag = 1; puts ("True "); break;} else flag = 0;} if (flag) break; if (! Flag) {continue ;}}if (I = len1) puts ("False");} return 0 ;}

At first, I wanted to copy other people's code. Later I was told by senior students to do it by myself, so I was able to do it by myself. Only in this way can I continue to improve, come on!