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 <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;
}