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 FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
long long n,s;
long long check(long long b,long long n){
     long long ans=0;
     while(n!=0){
        ans+=n%b;
        n/=b;
     }
     return ans;
}
int main(){
	cin>>n>>s;
		     //cout<<n<<s;

	long long ans=-1;
	REP(i,sqrt(n)+2){
	    if(i==0||i==1)continue;
		       //cout<<i<<" "<<check(i,n)<<endl;
	    if(check(i,n)==s){
	       ans=i;
	       break;
	    }
	}

        if(ans==-1&&s==n){
           ans=n+1;
        }
        //if(ans==-1){
           REP(i,sqrt(n)+2){
               if(i==0)continue;
               if((n-s)%i==0){
                  long long b=(n-s)/i+1;
                  if(b<2)continue;
                  long long q=n%b;
                  if(i+q==s){
                     if(ans==-1)ans=b;
                     ans=min(b,ans);
                  }
               }
           }
        //}

        cout<<ans<<endl;
	return(0);
}