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 ll long long
#define inf 0x3f3f3f3f3f3f3f3fLL
#define rep(i,a,b) for(register ll i=(a);i<=(b);i++)
#define dep(i,a,b) for(register ll i=(a);i>=(b);i--)
using namespace std;
const int maxn=50+5;
const int maxm=5000+5;
//const double pi=acos(-1.0);
//const double eps=1e-9;
const ll mo=1e9+7;
ll n,m,k;
int a[maxn];
ll ans,tmp,cnt;
char s[maxn];
ll power(ll a,ll n)
{
    ll sum=1;
    while(n)
    {
        if(n&1) sum=sum*a%mo;
        n>>=1;
        a=a*a%mo;
    }
    return sum;
}
ll f(ll b,ll n)
{
    if(n<b) return n;
    return n%b+f(b,n/b);
}
int main()
{
    while(scanf("%lld%lld",&n,&m)!=EOF)
    {
        ll ans=inf,tmp=sqrt(n*1.0);
        for(ll i=1;i<=tmp;i++)
        {
            ll b=n/i;
            ll x=(n/i)-n/(i+1);
            if(m-i<x&&m-i>=0) ans=min(ans,b);
        }
        for(ll i=2;i<=tmp;i++)
        {
            if(f(i,n)==m)
            {ans=min(i,ans);break;}
        }
        if(ans==inf) ans=-1;
        printf("%lld\n",ans);
    }
    return 0;
}