dpal.c
· 1.5 KiB · C
Ham
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char* rev(char buf[]) {
for (int i = 0; i < strlen(buf); i++) {
char t = buf[i];
buf[i] = buf[strlen(buf) - i - 1];
buf[strlen(buf) - i - 1] = t;
}
return buf;
}
char* convertToBase(int number, int base, char buf[]) {
int idx = 0;
char c[2];
while (number > 0) {
sprintf(c, "%d", number % base);
// if we were converting the base of anything bigger than
// base10, we would have needed a special function to return
// A - F for hexadecimal formats.
buf[idx++] = c[0];
number /= base;
}
buf[idx] = '\0';
// we'll have to reverse the string before returning it,
// because we're dividing the original number from the "one's" place -- remainder of
// number % base is always the least significant bits
return rev(buf);
}
int isPalindrome(char* str) {
int ptra = 0;
int ptrb = strlen(str) - 1;
while (ptra <= ptrb) {
if (str[ptra] != str[ptrb]) {
return 0;
}
ptra++;
ptrb--;
}
return 1;
}
int main() {
int S = 25;
int N = 3;
int len = N;
int* palindromeNums = (int*)malloc(len * sizeof(int));
int idx = 0;
char buf[100];
while (N >= 0) {
int cnt = 0;
for (int i = 2; i < 11; i++) {
char* retStr = convertToBase(S, i, buf);
if (isPalindrome(retStr)) {
cnt++;
}
if (cnt > 1) {
palindromeNums[idx++] = S;
break;
}
}
S++;
N--;
}
for (int i = 0; i < len; i++) {
printf("%d\n", palindromeNums[i]);
}
}
| 1 | #include <stdio.h> |
| 2 | #include <string.h> |
| 3 | #include <stdlib.h> |
| 4 | |
| 5 | char* rev(char buf[]) { |
| 6 | for (int i = 0; i < strlen(buf); i++) { |
| 7 | char t = buf[i]; |
| 8 | buf[i] = buf[strlen(buf) - i - 1]; |
| 9 | buf[strlen(buf) - i - 1] = t; |
| 10 | } |
| 11 | return buf; |
| 12 | } |
| 13 | |
| 14 | char* convertToBase(int number, int base, char buf[]) { |
| 15 | int idx = 0; |
| 16 | char c[2]; |
| 17 | while (number > 0) { |
| 18 | sprintf(c, "%d", number % base); |
| 19 | // if we were converting the base of anything bigger than |
| 20 | // base10, we would have needed a special function to return |
| 21 | // A - F for hexadecimal formats. |
| 22 | buf[idx++] = c[0]; |
| 23 | number /= base; |
| 24 | } |
| 25 | buf[idx] = '\0'; |
| 26 | // we'll have to reverse the string before returning it, |
| 27 | // because we're dividing the original number from the "one's" place -- remainder of |
| 28 | // number % base is always the least significant bits |
| 29 | return rev(buf); |
| 30 | } |
| 31 | |
| 32 | int isPalindrome(char* str) { |
| 33 | int ptra = 0; |
| 34 | int ptrb = strlen(str) - 1; |
| 35 | while (ptra <= ptrb) { |
| 36 | if (str[ptra] != str[ptrb]) { |
| 37 | return 0; |
| 38 | } |
| 39 | ptra++; |
| 40 | ptrb--; |
| 41 | } |
| 42 | return 1; |
| 43 | } |
| 44 | |
| 45 | int main() { |
| 46 | int S = 25; |
| 47 | int N = 3; |
| 48 | |
| 49 | int len = N; |
| 50 | int* palindromeNums = (int*)malloc(len * sizeof(int)); |
| 51 | int idx = 0; |
| 52 | char buf[100]; |
| 53 | |
| 54 | while (N >= 0) { |
| 55 | int cnt = 0; |
| 56 | for (int i = 2; i < 11; i++) { |
| 57 | char* retStr = convertToBase(S, i, buf); |
| 58 | if (isPalindrome(retStr)) { |
| 59 | cnt++; |
| 60 | } |
| 61 | if (cnt > 1) { |
| 62 | palindromeNums[idx++] = S; |
| 63 | break; |
| 64 | } |
| 65 | } |
| 66 | S++; |
| 67 | N--; |
| 68 | } |
| 69 | |
| 70 | for (int i = 0; i < len; i++) { |
| 71 | printf("%d\n", palindromeNums[i]); |
| 72 | } |
| 73 | |
| 74 | } |