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>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define LL long long
#define st first
#define nd second
using namespace std;
LL f(LL x,LL y){
    if(x<y)
        return x;
    return f(x/y,y)+x%y;
}
LL find(LL x,LL y){
    LL ret=1e18;
    LL f=1,l=1e18;
    while(f<=l){
        LL md=(l+f)/2;
        //cout<<x<<" "<<y<<" "<<l<<" "<<f<<endl;
        if(x/md>y)
            f=md+1;
        else if(x/md==y){
            ret=md;
            l=md-1;
        }
        else
            l=md-1;
        if(f==l)
            break;
    }
    if(x/f==y&&ret>f)
        return f;

    return ret;
}
int main(){
    LL n,s;
    cin>>n>>s;
    LL nw=sqrt(n);
    for(int i=2;i<=nw;++i){
        if(f(n,i)==s){
            cout<<i<<endl;
            return 0;
        }
    }
    for(int i=nw-1;i>=1;--i){
        LL base=find(n,i),add=n%base;
        //cout<<i<<" "<<base<<" "<<add<<" "<<s-i<<endl;
        if(s-i>add)
            base+=s-i-add;
        if(n/base==i&&n%base==s-i){
            cout<<base<<endl;
            return 0;
        }
    }
    if(s==n) {
        cout<<s+1<<endl;
        return 0;
    }
    puts("-1");
}