carloscn
commented
1 year ago Analysis
int32_t count_even(int32_t num)
{
int32_t ret = 0;
if (0 == num) {
return ret;
}
for (int32_t i = 1; i <= num; i ++) {
int32_t e = 0;
int32_t s = 0;
int32_t j = i;
while (j != 0) {
e = j % 10;
j = j / 10;
s += e;
}
if (!(s & 1)) {
ret += 1;
}
}
return ret;
}
Description
Given a positive integer num, return the number of positive integers less than or equal to num whose digit sums are even.
The digit sum of a positive integer is the sum of all its digits.
Example 1:
Input: num = 4 Output: 2 Explanation: The only integers less than or equal to 4 whose digit sums are even are 2 and 4.
Example 2:
Input: num = 30 Output: 14 Explanation: The 14 integers less than or equal to 30 whose digit sums are even are 2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28.
Constraints:
1 <= num <= 1000