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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define all(x) (x).begin(), (x).end()

typedef long long ll;

ll n, s;
ll ans = 1e18;
const int N = 317000;

void check(ll b) {
  ll v = n, sum = 0;
  while (v) {
    sum += v % b; v /= b;
  }
  if (sum == s) {
    ans = min(ans, b);
  }
}

void relax(ll a) {
  ll b = s - a;
  ll B = (n - s) / a + 1;
  if (a < B && b < B && 0 <= b && 2 <= B && a * B + b == n) {
    ans = min(ans, B);
  }
}

void twoDigits() {
  if (n - s <= 0) return;
  ll v = n - s;
  for (ll mul = 1; mul * mul <= v; mul++) {
     if (v % mul == 0) {
       relax(mul);
       relax(v / mul);
     }
  }
}

void solve() {
  cin >> n >> s;
  if (s == 1) {
    ans = n;
  }
  if (n == s) {
    ans = min(ans, n + 1);
  }
  for (int b = 2; b < N; b++) {
    check(b);
  }
  twoDigits();
  cout << (ans == 1e18 ? -1 : ans);
}

// CHECK LIMITS (n <= 10^5)
// CHECK CORNER CASES (n == 1)

int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  //cout << 1.0 * clock() / CLOCKS_PER_SEC << endl;
  solve();
}