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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UncheckedIOException;
import java.util.StringTokenizer;

class Main{
	public static void main(String[] args) {
		SC sc=new SC(System.in);
		long n=sc.nextLong();
		long s=sc.nextLong();
		if(n==s) {
			pl(s+1);
			exit();
		}
		long sq=root(n);
		long sum=Long.MAX_VALUE;
		for(long i=2; i<=sq; i++) {	//k　進数で表す　2 < n < sqrt(n)　までを調べると桁数が3いじょうになる
			sum=ketawa(n,i);
			if(sum==s) {
				sum=Math.min(sum,i);
			}
		}
		for(long i=sq; i>=1; i--) {
			long ato=s-i;		//
			if((n-ato)%i==0) {
				sum=Math.min(sum, (n-ato)/i);
				exit();
			}
		}
		if(sum<Long.MAX_VALUE) {
			pl(sum);
		}
		else {
			pl(-1);
		}
	}
	static class SC {
		private BufferedReader reader = null;
		private StringTokenizer tokenizer = null;
		public SC(InputStream in) {
			reader = new BufferedReader(new InputStreamReader(in));
		}
		public String next() {
			if (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new UncheckedIOException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public double nextFloat() {
			return Float.parseFloat(next());
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new UncheckedIOException(e);
			}
		}
	}
	public static void pl(Object o) {
		System.out.println(o);
	}
	public static void p(Object o) {
		System.out.print(o);
	}
	public static void exit() {
		System.exit(0);
	}
	static long root(long a) {

		int ketasu=0;
		long tmp=a;
		while(tmp>0) {
			ketasu++;
			tmp/=10;
		}
		int constant=(ketasu+1)/2;	//よく使うので定数化
		if(ketasu>=1) {
			long[] suuji=new long[constant];
			tmp=a;
			for(int i=0; i<constant; i++) {
				suuji[constant-1-i]=a%100;
				a/=100;
			}
			long ans=0;
			long kai=0;
			long mae=0;
			if(constant<=1) {
				for(int i=10; i>=0; i--) {
					if(suuji[0]>=i*i) {
						return i;
					}
				}
			}
			else {
				for(int i=0; i<constant; i++) {
					mae+=kai*2;
					for(int j=9; j>=0; j--) {
						if((mae*10+j)*j<=suuji[i]) {
							ans=ans*10+j;
							mae*=10;
							kai=j;
							if(i+1<constant) {
								suuji[i+1]+=(suuji[i]-(mae+j)*j)*100;
							}
							break;
						}
					}
				}
			}
			return ans;
		}
		else {
			return 0;
		}
	}
	public static long ketawa(long a, long tei) {
		long sum=0;
		while(a>0) {
			sum+=a%tei;
			a/=tei;
		}
		return sum;
	}
}