2014-2015 ACM-ICPC, NEERC, Moscow Subregional Contest —— E. Equal Digits

E. Equal Digits

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

For the given integer N and digit D, find the minimal integer K ≥ 2 such that the representation ofN in the positional numeral system with baseK contains the maximum possible consecutive number of digitsD at the end.

Input

The input contains two integers N and D (0 ≤ N ≤ 10¹⁵,0 ≤ D ≤ 9).

Output

Output two integers: K, the answer to the problem, andR, the the number of consecutive digits D at the end of the representation of N in the positional numeral system with baseK.

Sample test(s)

Input

3 1

Output

2 2

Input

29 9

Output

10 1

Input

0 4

Output

2 0

Input

90 1

Output

89 2

                                                               

题意：

给出n 和 d , 要求得到用k 进制得到n，最后一个数要求为d，求出从后数d 的数量，要求数量最多是k 尽可能小。

思路：

要使n - d 用k 进制表示出，所以求出n-d 的因子，暴力枚举得到最大答案。

特殊情况是：n== d 和n< d

CODE:

#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
vector v;
ll n, d;
void work(ll nn)
{
    for(ll i = 1; i * i <= nn; ++i){
        if(nn%i == 0){
        if(i > d)
            v.push_back(i);
        if((nn/i) > d && i*i != nn)
            v.push_back(nn/i);
        }
    }
}
int main()
{
    //freopen("in", "r", stdin);
    while(~scanf("%I64d %I64d", &n, &d)){
        if(n == d){
            if(n <= 1) printf("2 1\n");
            else printf("%I64d 1\n",n + 1);
            continue;
        }
        if(n < d){
            printf("2 0\n");
            continue;
        }

        v.clear();
        work(n - d);
        ll r = 0, k = 2;
        for(ll i = 0; i < v.size(); ++i){
            ll n1 = n, s = 0;
            if(v[i] > 1){
                while(n1){
                    if(n1 % v[i] != d) break;
                    else{
                        s++;
                        n1 = n1/v[i];
                    }
                }
                if(s > r){
                    r = s;
                    k = v[i];
                }
                if(s == r && v[i] < k){
                    k = v[i];
                }
            }
        }
        printf("%I64d %I64d\n", k, r);
    }
    return 0;
}