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 <map>
#include <algorithm>
#include <cassert>
#include <climits>
#include <complex>
#include <cstdio>
#include <string>
#include <iostream>
#include <queue>
#include <string>
#include <tuple>
#include <vector>
using namespace std;
using ll = long long;
using ld = long double;
template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
#define DEBUG(X) cerr<<"  "<<#X<<" = "<<X<<endl;
#define DUMP(A, n) for (auto x=begin(A); x!=begin(A)+n;x++){cout <<*x<< ' ';} cout<<endl;
#define DUMPP(A, n, m) for (auto x=begin(A); x != begin(A)+n;x++) {for (auto y=begin(*x); y != begin(*x)+m;)cout <<*y++<< ' '; cout<<endl;};
#define DUMPM(M) for (auto itr=mp.begin(); itr!=mp.end(); itr++) {cout<<itr->first<<" -> "<<itr->second<<endl;}
#define FOR(i,x,y) for(int i=(x);i<(int)(y);i++)
#define FORP(i,x,y) for(int i=(x);i<=(int)(y);i++)
#define REP(i,y) for(int i=0;i<(int)(y);i++)
#define REPP(i,y) for(int i=1;i<=(int)(y);i++)
#define RREP(i,n) for(int i=(int)(n)-1;i>=0;i--)

ll n,s;

int f(ll b, ll n)
{
  if (n >= b)
    return (int)(n%b) + f(b, (ll)(n/b));
  else
    return (int)n;
}

int main() {
  cin >> n >> s;
  ll nr = sqrt(n);
  for(ll b=2;b<nr;b++)
  {
    int sum = f(b, n);
    if (sum==s) {
      cout << b << endl;
      return 0;
    }
  }

  // 一桁
  if (s==n)
  {
    cout << n+1 << endl;
    return 0;
  }

  if (s>18) {
    cout << "-1" << endl;
    return 0;
  }
  for(int p=1;p<=9;p++)
  {
    ll q = s-p;
    if ((n-q)%p==0&&10>q&&q>=0) {
      ll b = (n-q)/p;

      cout << b << endl;
      return 0;
    }
  }
  cout << "-1" << endl;
  return 0;
}