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 digitsD 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 <iostream>#include <cstdio>#include <vector>#include <cmath>using namespace std;typedef long long ll;vector<ll> 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;}