Error while using pow() in C++

Question

I was trying to solve a function problem in which I had to count the number of times digit 1 appeared in all non-negative integers less than n(given).

Here's my code:

int ones(int n, int d)
{
    int rounddown = n - n % pow(10, d+1);
    int roundup = rounddown + pow(10, d+1);
    int right = n % pow(10, d);
    int dig = (n/pow(10, d)) % 10;
    if(dig<1)
        return rounddown/10;
    else if(dig==1)
        return rounddown/10 + right + 1;
    else
        return roundup/10;
}

int countDigitOne(int n) {

    int count = 0;
    string s = to_string(n);
    for(int i=0;i<s.length();i++)
    count+=ones(n, i);   
    return count;        
}

But the following compilation error shows up:

Line 3: invalid operands of types '__gnu_cxx::__promote_2::__type {aka double}' and 'int' to binary 'operator%'

Answer 1

The main problem is type conversion. The result of pow is double. The modulo operator does not work for double. You need to take fmod.

Fix your line 3 like this:

int rounddown = (int)(n - fmod(n, pow(10, d +1));

Because your values are all in the domain of integer you can also use:

int rounddown = n - n % (int)(pow(10, d + 1));

as suggested by others.

---

Just for completeness ... If you are not forced to use an arithmetic approach you can just char compare:

#include<iostream>
#include<string>

using namespace std;

int countOnesInNumber(string s)
{
    int res = 0;
    for(int i = 0; i < s.length(); i++)
    {
        if(s[i] == '1')
        {
            res++;
        }
    }
    return res;
}

long countOnes(int upper)
{
    long result = 0;
    for(int i = 0; i < upper; i++)
    {
        result += countOnesInNumber(std::to_string(i));
    }
    return result;
}

int main()
{
    string in;
    cout << "Please enter a number:";
    cin >> in;
    cout << endl;
    cout << "you choose:" << in << endl;
    int n = stoi(in);
    cout << "there are " << countOnes(n) << " ones under " << n << endl;

    cin.ignore();
    cin.get();
}

---

There is also a more sophisticated approach. The digits repeat for every magnitude. There is 1 one under 10, there are 10 times 1 one under 100 plus another 10 ones for the number 10...19. And so on. You can just calculate the number of ones under a given magnitude with:

int exp = (int)log10(n);
int ones = exp * (int)pow(10, exp - 1);

where n must be a magnitude of 10 (eg. 10, 100, 1000, 10000 ...). If you are good at maths you might even find a complete closed formula.