CF#538（div 2） C. Trailing Loves (or L'oeufs?) 【经典数论 n！的素因子分解】-优快云博客

任意门：http://codeforces.com/contest/1114/problem/C

The number "zero" is called "love" (or "l'oeuf" to be precise, literally means "egg" in French), for example when denoting the zero score in a game of tennis.

Aki is fond of numbers, especially those with trailing zeros. For example, the number

However, Aki believes, that the number of trailing zeros of a number is not static, but depends on the base (radix) it is represented in. Thus, he considers a few scenarios with some numbers and bases. And now, since the numbers he used become quite bizarre, he asks you to help him to calculate the beauty of these numbers.

Given two integers

Input

The only line of the input contains two integers

Output

Print an only integer — the number of trailing zero digits in the

Examples

input

Copy

6 9

output

Copy

input

Copy

38 11

output

Copy

input

Copy

5 2

output

Copy

input

Copy

5 10

output

Copy

Note

In the first example,

In the third and fourth example,

The representation of the number

You can read more about bases here.

题意概括：

求 n！转化为 b 进制下末尾有多少个 0.

解题思路：

原题：swjtuOJ 2090

这是一道很好的数论题。

首先转换一下思路：

要求 n! 在 b 进制下有多少个尾 0 就相当于求 n! % (b^k) == 0 的最大 k。

那么我们现在把 n! 看作一个数 A。问题就是求 A % (b^k) == 0 的最大 k;

我们知道有素数分解定理： b = p1^a1 * p2^a2 * p3^a3 ...；

那么我们如果可以求得 A 里面 p1^b1 * p2^b2 * p3^b3 ... 的 b1, b2, b3...

那么答案 ans = min(ans, ai/bi ) 了也就是要整除，首先要满足最小的那个能整除。

(1)首先对 b 进行素因子分解，直接暴力（log b）, 用一个数组离散化形成该素因子的编号和该素因子的幂的映射或者用map存储该素因子的幂，得到所有素因子以及素因子的幂

(2)对于每一个素因子p，计算对应的 A（即 n! ) 中素因子p的幂，两者相除取所有p幂的最小值就是对应的最大整数。

这里求 n! 下素因子 p 的幂用累除法，因为存在推论：

n！下 p 的幂 = [ n/p ] + [ n/(p^2) ] + [ n/(p^3) ] ...

学习：

https://blog.youkuaiyun.com/Wen_Yongqi/article/details/86976902

AC code：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<vector>
 6 #include<queue>
 7 #include<cmath>
 8 #include<set>
 9 #include<map>
10 #define INF 0x3f3f3f3f
11 #define LL long long
12 using namespace std;
13 const int MAXN = 2e6+1000;
14 LL N, B;
15 vector<LL>prime;
16 //LL num[MAXN];
17 map<LL, int>mmp;
18 void get_p(LL n)
19 {
20     LL len = sqrt((double)n);
21     //printf("len %lld\n", len);
22     for(LL i = 2; i <= len; i++){
23         if(n%i == 0){
24             prime.push_back(i);
25             while(n%i == 0){
26                 n/=i;
27                 //num[prime.size()-1]++;
28                 mmp[i]++;
29             }
30         }
31     }
32     if(n != 1){
33         prime.push_back(n);
34 //        num[prime.size()-1]++;
35         mmp[n]++;
36     }
37 }
38 
39 LL calc(LL n, LL p)
40 {
41     LL res = 0;
42     while(n){
43         res += n/p;
44         n/=p;
45         //puts("zjy");
46     }
47     return res;
48 }
49 
50 int main()
51 {
52     LL sum = 1LL;
53     scanf("%I64d %I64d", &N, &B);
54     get_p(B);
55     LL ans = (1LL<<62);
56     for(LL i = 0; i < prime.size(); i++){
57 //        ans = min(ans, calc(N, prime[i])/num[i]);
58         //puts("zjy");
59         ans = min(ans, calc(N, prime[i])/mmp[prime[i]]);
60     }
61 
62     printf("%I64d\n", ans);
63     return 0;
64 }