Kyselyt

/*
	Written by, Tuukka Yildirim.
	
	input: q -> 1 <= q <= 1000
		   q lines, where 1 <= k <= 10**18 (** denotes exponentiation).
	In other words, maximum value of k is: 1000000000000000000.
	
	I generated a lookup table <tenPowerIndexes>. The table contains data
	that yields how many digits are in the range of <k>. For example,
	10 -> at this point, numbers start to contain 2 digits.
	190 -> at this point, numbers start to contain 3 digits.
	... and so on. This array can be generated with the function <generatePowerIndexes()>.
	For arbitrary sized integers, the function could be trivially modified to work for up to infinity.
	<howManyDigits(k)> fetches this data, based on <k>.
	
	The idea is to find, which power of 10 digit <k> is pointing to, and subsequently calculate it.
	By following the comments in the function <findDigit(k)> and analyzing the math, you'll get the picture.
*/


#include <iostream>
#include <string>

using namespace std;
typedef long long ll;
ll tenPowerIndexes[17] = {10LL, 190LL, 2890LL, 38890LL, 488890LL,
						5888890LL, 68888890LL, 788888890LL,
						8888888890LL, 98888888890LL, 1088888888890LL, 11888888888890LL,
						128888888888890LL, 1388888888888890LL, 14888888888888890LL,
						158888888888888890LL, 1688888888888888890LL};
			
ll customPow(ll base, ll exponent)
{
	if (exponent == 0LL) return 1LL;
	ll res = base;
	while (exponent != 1LL)
	{
		res *= base;
		exponent--;
	}
	return res;
}
			
// This is a function that could generate the list <tenPowerIndexes>.
void generatePowerIndexes()
{
	ll prevSum = 0LL;
	
	for (ll i = 0LL; i < 17LL; i++)
	{
		ll range = 9LL * customPow(10LL, i);
		ll digCount = i + 1LL;
		
		ll newSum = prevSum + (range * digCount);
		prevSum = newSum;
		tenPowerIndexes[i] = newSum + 1;
	}
}

ll howManyDigits(ll k)
{
	ll digits = 1LL;
	for (int i = 0; i < 17; i++)
	{
		if (k < tenPowerIndexes[i]) break;
		else digits++;
	}
	
	return digits;
}

ll findDigit(ll k)
{
	if (k < 10LL) return k;
	
	ll d = howManyDigits(k);
	// d-space is now calculated.
	
	// Normalized index starting from the d-digit number.
	// ... 97 98 99 100 101 102 ...
	// For example, 100 will have the index 0 ==> p.
	ll p = k - tenPowerIndexes[d - 2];
	
	
	// Let's calculate what power of 10 digit are we trying to find (10**0, 10**1 etc).
	ll region = (d - 1) - (p % d);
	
	if (region == (d - 1))
	{
		// +1 required.
		return (p / (d * customPow(10LL, d - 1))) + 1;
	}
	
	return (p / (d * customPow(10LL, region))) % 10;
}

int main() 
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	
	int q;
	cin >> q;
	cin.ignore();
	
	for (int i = 0; i < q; i++)
	{
		ll search;
		cin >> search;
		cin.ignore();
		cout << findDigit(search) << "\n";
	}
	
	return 0;
}

Test 1

Group: 1

Verdict: ACCEPTED

Test 2

Group: 2

Verdict: ACCEPTED

Test 3

Group: 3

Verdict: ACCEPTED