// Solution to Project Euler problem #36
// palindromic in base10 and base2
#include <stdio.h>
int rev10(int num);
int rev2(int num);
main() {
int sum=0, i;
for (i=1; i<1000000; i+=2) { // skip evens since they have a leading zero in at least binary
if (i==rev10(i) && i == rev2(i)) {
sum += i;
printf("New solution: %d on %d\n", sum, i);
}
}
}
int rev10(int num) {
int newn=0;
while (num) {
newn *= 10; // shift already reveresed digits
newn += num%10; // put new digit on end
num /=10; // pull left digits in
}
return newn;
}
int rev2(int num) {
int newn = 0;
while (num) {
newn *= 2; // shift already reversed digits
newn += num & 1; // put new digit on end
num /= 2; // pull left digits in
}
return newn;
}