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 of N in the positional numeral system with base K contains the maximum possible consecutive number of digits D 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, and R, the the number of consecutive digits D at the end of the representation of Nin the positional numeral system with base K.

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，d，找出一个k，是n在k进制下，尾部的d的数量最大。

分析：

显然答案是一个正整数，因为我们可以取k=n-d进制。

那么，因为至少为1，则k必定整除于n-d

那么将n-d分解，将其因数全部拿出来。

暴力更新答案即可。

因数个数在根号级别，暴力计算复杂度在log级别，不会超时。

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <ctime>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <vector>
 #include <deque>
 #include <queue>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define Rep(i, n) for(int i = (0); i < (n); i++)
 #define Repn(i, n) for(int i = (n)-1; i >= 0; i--)
 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 #define Ford(i, t, s) for(int i = (t); i >= (s); i--)
 #define rep(i, s, t) for(int i = (s); i < (t); i++)
 #define repn(i, s, t) for(int i = (s)-1; i >= (t); i--)
 #define MIT (2147483647)
 #define MLL (1000000000000000000LL)
 #define INF (1000000001)
 #define mk make_pair
 #define ft first
 #define sd second
 #define clr(x, y) (memset(x, y, sizeof(x)))
 #define sqr(x) ((x)*(x))
 #define sz(x) ((int) (x).size())
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back

 template<class T>
 inline T Getint()
 {
     char Ch = ' ';
     T Ret = ;
     while(!(Ch >= '' && Ch <= '')) Ch = getchar();
     while(Ch >= '' && Ch <= '')
     {
         Ret = Ret *  + Ch - '';
         Ch = getchar();
     }
     return Ret;
 }

 LL n;
 int d;
 vector<LL> Factor;

 inline void Input()
 {
     cin >> n >> d;
 }

 inline void Solve()
 {
     if(n == d)
     {
         if(n <= ) puts("2 1");
         else cout << n + 1LL << "" << endl;
         return;
     }
     if(n < d)
     {
         puts("2 0");
         return;
     }

     LL T = n - d;
     for(LL i = ; i * i <= T; i++)
         if(T % i == )
         {
             if(i > d) Factor.pub(i);
             if(T / i > d) Factor.pub(T / i);
         }

     LL R = , k = ;
     int Length = sz(Factor), Cnt;
     Rep(i, Length)
     {
         if(Factor[i] == ) continue;

         T = n, Cnt = ;
         while(T % Factor[i] == d)
             T /= Factor[i], Cnt++;

         if(Cnt > R) R = Cnt, k = Factor[i];
         else if(Cnt == R) k = min(k, Factor[i]);
     }

     cout << k << ' ' << R << endl;
 }

 int main() {
     //freopen("E.in", "r", stdin);
      Input();
      Solve();
     return ;
 }