Checking if a number is a Simber - c++

Question

A simber is defined as a positive integer in which any odd digit, if present, occurs an odd number of times, 1414414 is a simber. 4 is even and it appears four time and 1 is odd and it appears 3 times.
Here are my variables:
int is_simber(int n)
{
int numberOfTimes = 0, length = 1, x = n;
bool answer;
vector <int> nmbrs = vector <int>();
//get how many digits are in the integer
do
{
x /= 10;
length++;
}
while(x != 0)
//get the digits in the integer
for(int i = 0; i<length; i++)
{
nmbrs.push_back(((n/10^i) % 10);
}
//checking how many times a digit occurs and also testing to see if the digits
//meet the requirements
for(int i = 0; i<length; i++)
{
for(int j = 0; j<length; j++)
{
if (nmbrs.at(i) == nmbrs.at(j))
{
numberOfTimes++;
}
}
if (nmbrs.at(i) % 2 == 0 && numberOfTimes % 2 == 0)
{
answer = true;
}
else if(nmbrs.at(i) % 2 == 1 && numberOfTimes % 2 == 1)
{
answer = true;
}
else if(nmbrs.at(i) % 2 == 0 && numberOfTimes % 2 == 1)
{
answer = false;
break;
}
else if(nmbrs.at(i) % 2 == 1 && numberOfTimes % 2 == 0)
{
answer = false;
break;
}
}
return answer;
}

Answer 1

Your code has compilation errors.
^ is Binary XOR Operator. You can't expect it to produce pow(10,i). So replace nmbrs.push_back(((n/10^i) % 10); with nmbrs.push_back(((n/pow(10,i)) % 10);
When I was able to remove the compilation errors, i realized your code has logical errors too. As your given definition of simber, int is_simber(int n) should be fair and simple.
bool is_simber( int n )
{
if ( n<0 ) return false; // simber is a positive integer
int digitCount[10] = {}; // initializing all with 0;
// digitCount array holds the number of occurance of a digit
// so d[1] holds the number of times 1 occurred
// d[2] holds the number of times 2 occurred and so on ...
while( n ){
int d = n%10;
n /= 10;
digitCount[d]++;
}
// we just have to check
// any odd digit, if present, occurs an odd number of times
for( int i=1; i<=9; i= i+2) // Attention i=i+2, to iterate over just odd numbers
{
if( digitCount[i] != 0 && digitCount[i]%2 == 0 ) return false;
}
return true;
}

Answer 2

My definition of is_simber
inline bool is_even(int n) { return n % 2 == 0; }
bool is_simber(int n) {
if (n < 0) return false;
int digits[10] = {0};
for (; n; n /= 10) ++digits[n % 10];
for (int i = 0; i < 10; i += 2)
if (!is_even(digits[i]) && is_even(digits[i + 1])) return false;
return true;
}
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