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 (GCC 5.4.1) (judge verdict: Wrong Answer)


#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

int main(void)
{
	//変数の宣言
	long long int n,s;
	static long long int num[1000010];
	long long int sqrtn;
	
	//よく使う変数
	int i;
	long long int j,k,l;
	int flag=0;
	int ans=0;
	int count=0;
	long long int temp,temp1,temp2;
	int max,min;
	int len;
	//データの読み込み
	
	scanf("%lld",&n);
	scanf("%lld",&s);
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
//	printf("n=%lld\n",n);

	if(n<s){
		printf("-1");
		return 0;
	}

	for(j=1;j<n;j++){
//		printf("i=%lld, i*i=%lld\n",i,i*i);
		if(j*j>=n){
			sqrtn=j;
			break;
		}
	}

//	printf("sqrtn=%lld,sqrtn^2=%lld,\n",sqrtn,sqrtn*sqrtn);

	for(i=1;i<sqrtn;i++){
//		printf("i=%d (n-s)%%i=%lld\n",i,(n-s)%i);
		if((n-s)%i==0){
			num[count]=i;
//			printf("num[%d]=%lld\n",count,num[count]);
			count++;
		}
	}
	
	for(i=count;i<2*count;i++){
		num[i]=(n-s)/num[2*count-1-i];
	}
	
	for(i=0;i<2*count;i++){
//		printf("num[%d]=%lld\n",i,num[i]);
		temp=n;
		temp2=0;
		while(temp!=0){
			temp2=temp2+temp%(num[i]+1);
			temp=temp/(num[i]+1);
//			printf("temp=%lld,temp2=%lld\n",temp,temp2);
		}
		if(temp2==s){
			printf("%lld",num[i]+1);
			return 0;
		}
	}
	
//	printf("計算部分終了\n");
	//出力
	
	printf("-1");
	
//	printf("結果の出力終了\n");
	
	return 0;
	
}