Codeforces 837D Round Subset - 动态规划 - 数论

Let's call the roundness of the number the number of zeros to which it ends.

You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible.

Input

The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).

The second line contains n space-separated integer numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10¹⁸).

Output

Print maximal roundness of product of the chosen subset of length k.

Examples

Input

3 2
50 4 20

Output

3

Input

5 3
15 16 3 25 9

Output

3

Input

3 3
9 77 13

Output

0

Note

In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.

In the second example subset [15, 16, 25] has product 6000, roundness 3.

In the third example all subsets has product with roundness 0.

---

　　题目大意 给定一个数组，从中选出k个数(不能选同一个数)，使得这k个数的乘积的末尾的零的个数最多。

　　根据小学的数学知识，我们知道一个数的末尾有多个零取决于它质因数分解后2的指数和5的指数的最小值。(10 = 2 × 5)

　　所以我们初步得到动态规划的状态f[i][j][k]表示，从前i个数中，选出j个数，使得它们的乘积质因数分解后2的指数为k，5的指数最大为多少。

　　显然它有两种转移:选第i + 1个数，不选第i + 1个数。所以转移是显然的。

　　然后您会得到MLE，所以加上黑科技bfs版 + 滚动数组动态规划，不知道能不能卡过，但是有一种很简单的优化方法。

　　显然将题目中输入的一个数质因数分解后5的指数不会超过。所以我们可以把5个个数作为状态，2的个数作为状态的值，这样总状态数不会超过200³ * 25(刚刚是乘64)

　　所以继续黑科技优化内存和时间就过了。

Code

 /**
  * Codeforces
  * Problem#837D
  * Accepted
  * Time: 187ms
  * Memory: 10604k
  */
 #include <bits/stdc++.h>
 using namespace std;
 typedef bool boolean;
 #define smax(a, b) a = max(a, b);
 template<typename T>
 inline boolean readInteger(T& u){
     char x;
     int aFlag = ;
     while(!isdigit((x = getchar())) && x != '-' && x != -);
     if(x == -) {
         ungetc(x, stdin);
         return false;
     }
     if(x == '-'){
         x = getchar();
         aFlag = -;
     }
     for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
     ungetc(x, stdin);
     u *= aFlag;
     return true;
 }

 typedef class Status {
     public:
         int stage;
         int seced;
         int c5;
 }Status;

 int n, k;
 int *c2s, *c5s;
 int res = ;

 inline void init() {
     long long x;
     readInteger(n);
     readInteger(k);
     c2s = new int[(n + )];
     c5s = new int[(n + )];
     for(int i = ; i <= n; i++) {
         c2s[i] = c5s[i] = ;
         readInteger(x);
         while(!(x & ))    x >>= , c2s[i]++;
         while(!(x % ))    x /= , c5s[i]++;
     }
 }

 queue<Status> que;
 int f[][][];
 inline void dp() {
     int last = ;
     Status sta = (Status) {, , };
     que.push(sta);
     memset(f, -, sizeof(f));
     f[][][] = ;
     while(!que.empty()) {
         Status e = que.front();
         que.pop();

         int lastf = f[e.stage & ][e.seced][e.c5];
         Status eu = e;
         eu.stage++;
         if(eu.stage != last) {
             last = eu.stage;
             memset(f[eu.stage & ], -, sizeof(f[]));
         }

         if(f[eu.stage & ][eu.seced][eu.c5] == - && eu.stage < n && eu.seced <= k)
             que.push(eu);
         else if(eu.stage == n && eu.seced == k)
             smax(res, min(eu.c5, lastf));
         smax(f[eu.stage & ][eu.seced][eu.c5], lastf);

         eu.seced++, eu.c5 += c5s[eu.stage];
         if(f[eu.stage & ][eu.seced][eu.c5] == - && eu.stage < n && eu.seced <= k)
             que.push(eu);
         else if(eu.stage == n && eu.seced == k)
             smax(res, min(eu.c5, lastf + c2s[eu.stage]));
         smax(f[eu.stage & ][eu.seced][eu.c5], lastf + c2s[eu.stage]);
     }
 }

 inline void solve() {
     printf("%d\n", res);
 }

 int main() {
     init();
     dp();
     solve();
     return ;
 }