Problem:
Score : 500 points

Problem Statement

For integers b (b \geq 2) and n (n \geq 1), let the function f(b,n) be defined as follows:

f(b,n) = n, when n < b

f(b,n) = f(b,\,{\rm floor}(n / b)) + (n \ {\rm mod} \ b), when n \geq b

Here, {\rm floor}(n / b) denotes the largest integer not exceeding n / b,
and n \ {\rm mod} \ b denotes the remainder of n divided by b.

Less formally, f(b,n) is equal to the sum of the digits of n written in base b.
For example, the following hold:

f(10,\,87654)=8+7+6+5+4=30

f(100,\,87654)=8+76+54=138

You are given integers n and s.
Determine if there exists an integer b (b \geq 2) such that f(b,n)=s.
If the answer is positive, also find the smallest such b.

Constraints

1 \leq n \leq 10^{11}

1 \leq s \leq 10^{11}

n,\,s are integers.

Input

The input is given from Standard Input in the following format:

n
s

Output

If there exists an integer b (b \geq 2) such that f(b,n)=s, print the smallest such b.
If such b does not exist, print -1 instead.

Sample Input 1

87654
30

Sample Output 1

10

Sample Input 2

87654
138

Sample Output 2

100

Sample Input 3

87654
45678

Sample Output 3

-1

Sample Input 4

31415926535
1

Sample Output 4

31415926535

Sample Input 5

1
31415926535

Sample Output 5

-1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)


#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{

	ios_base::sync_with_stdio(0);
	cout.tie(0);
	cin.tie(0);

// #ifndef ONLINE_JUDGE
// 	freopen("input.txt" , "r" , stdin);
// 	freopen("output.txt" , "w" , stdout);
// #endif

	ll n, s, f = 0;
	cin >> n >> s;
	for (ll i = 2; i * i <= n; i++)
	{
		ll y = n, ans = 0;
		while (y)
		{
			ll mod = y % i;
			ans += mod;
			y = y / i;
		}
		if (ans == s)
		{
			f = i;
			break;
		}
	}
	if (!f)
	{
		ll n1 = n - s;
		for (ll i = 1; i * i <= n1; i++)
		{
			if (n1 % i == 0)
			{
				ll b = n1 / i + 1;
				ll y = n, ans = 0;
				while (y)
				{
					ll mod = y % b;
					y = y / b;
					ans += mod;
				}
				if (ans == s)
					f = b;
			}
		}
	}
	if (!f)
		cout << -1 << endl;
	else
		cout << f << endl;



}