#include <stdio.h>
#include <string.h>
int is_mynumber(char a) {
int is_number = 0;
if (a >= '2' && a <= '9') {
is_number = 1;
}
return is_number;
}
void result(char a[]) {
int i = 0;
while (a[i] != '\0') {
if (a[i + 1] != '\0') {
i++;
int is_number = is_mynumber(a[i]);
if (is_number) {
int times = a[i] - '0';
while (times > 0) {
printf("%c", a[i - 1]);
times--;
}
}
if (!is_number) {
printf("%c", a[i - 1]);
}
i++;
} else if (a[i + 1] == '\0') {
printf("%c", a[i]);
return;
}
}
}
int main() {
char a[1001];
scanf("%[^\n]", a);
result(a);
return 0;
}
This code reads a string and tries to expand it by repeating characters based on digits that follow them. For example, "a2b3" would print "aabbb". The logic walks through the string, checks if the next character is a digit, and if so prints the previous character that many times.
- The loop increments `i` twice in some cases, which can cause you to skip characters or read past the end of the string. Look carefully at where `i` is increased and whether the condition `a[i+1] != '\0'` is checked at the right time.
- The function `is_mynumber` only treats '2' through '9' as digits, but '0' and '1' are also digits. Consider what should happen if a '1' or '0' appears after a character.