Kyselyt

#include <iostream>
#include <vector>
 
using namespace std;
typedef long long ll;
 
ll power(ll a, ll b) {
    if(b == 0) {
        return 1;
    }
    if(b%2 == 0) {
        ll c = power(a, b/2);
        c = c*c;
        return c;
    }
    ll c = power(a,b - 1);
    return a * c;
}
 
int main() {
    int q;
    cin >> q;
    for(int a = 1; a <= q; a++) {
        ll k;
        cin >> k;
        ll digitsCurrently = 1;
        ll amountOfNumbers = 9;
        ll totalAmountOfDigits = 9;
        //Determine the amount of digits in the number the k:th digit is in
        
        while(totalAmountOfDigits < k) {
            digitsCurrently++;
            amountOfNumbers *= 10;
            totalAmountOfDigits += digitsCurrently*amountOfNumbers;
        }
        ll smallerNumberDigits = totalAmountOfDigits - digitsCurrently*amountOfNumbers;
        
        //Stop caring about the small numbers before the numbers with correct amount of digits. Use zero-indexing.
        k = k - smallerNumberDigits - 1;
        //Determine the location of the k:th digit
        ll spotInNumber = k%digitsCurrently;
        ll theNumber = k/digitsCurrently + power(10, digitsCurrently - 1);
        //Break theNumber into digits
        vector<int> digitsOfTheNumber;
        while(theNumber > 0) {
            digitsOfTheNumber.push_back(theNumber%10);
            theNumber /= 10;
        }
        //Print the spotInNumber:th digit. Note that digitsOfTheNumber contains the digits in reverse
        cout << digitsOfTheNumber[digitsOfTheNumber.size() - (spotInNumber + 1)];
        
    }
}
 

Test 1

Group: 1

Verdict: WRONG ANSWER

Test 2

Group: 2

Verdict: WRONG ANSWER

Test 3

Group: 3

Verdict: WRONG ANSWER