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 REP(i,n) for(int i=0;i<n;i++)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define FOR(i,m,n) for(int i=m;i<n;i++)
#define EPS (1e-7)

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;

int vx[]={0,1,0,-1};
int vy[]={1,0,-1,0};

int dx[]={1,0,-1,1,-1,1,0,-1};
int dy[]={1,1,1,0,0,-1,-1,-1};

const ll mod=1e9+7;
const int INF=INT_MAX/2;
const int MAX=1e9;

ll n,s;
bool check(ll B, ll N, ll S){
	ll res=0;
	if(B>N){
		res=N;
	}else{
		while(N>0){
			res+=N%B;
			N/=B;
		}
	}
	if(res==S)return true;
	else return false;
}

int main(){
	cin>>n>>s;
	if(n<s){
		cout<<-1<<endl;
		return 0;
	}else if(n==s){
		cout<<n+1<<endl;
		return 0;
	}else{
		for(ll i=2;i<sqrt(n);i++){
			if(check(i,n,s)){
				cout<<i<<endl;
				return 0;
			}
		}
		for(ll i=sqrt(n);i>=1;i--){
			if((n-s)%i!=0)continue;
			ll b=(n-s)/i+1;
			if(check(b,n,s)){
				cout<<b<<endl;
				return 0;
			}
		}
	}
	cout<<-1<<endl;
	return 0;
}