Educational Codeforces Round 26 D. Round Subset 动态规划

Posted 2020-10-01 starry

D. Round Subset

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.

 

给你n个数，从中取k个数，求使得k个数相乘最多有多少个后缀0。

一个神奇的动态规划。dp[i][j]表示取i个数，在有j个5的情况下最多有多少个2

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <cmath>
 6 #define ll long long
 7 using namespace std;
 8 struct Nod {
 9     ll a;
10     int x,y;
11     Nod(){
12         a = x = y = 0;
13     }
14 }nod[220];
15 void fun(Nod *p) {
16     ll x = p->a;
17     while(x%2 == 0){
18         p->x++;
19         x/=2;
20     }
21     x = p->a;
22     while(x%5 == 0){
23         p->y++;
24         x/=5;
25     }
26 }
27 int dp[210][18*210];
28 bool vis[220];
29 int main() {
30     int n, k,MAXY=0;
31     cin >> n >> k;
32     for(int i = 1; i <= n; i ++)
33         cin >> nod[i].a, fun(&nod[i]),MAXY += nod[i].y;
34     memset(dp, -1, sizeof(dp));
35     // for(int i = 1; i <= n; i ++)printf("%d %d\n",nod[i].x,nod[i].y);
36     // printf("%d\n",MAXY);
37     dp[0][0] = 0;
38     for(int i = 1; i <= n; i ++) {
39         for(int j = k; j >= 1; j --) {
40             for(int k = MAXY; k >= 0; k --) {
41                 if(k-nod[i].y < 0) continue;
42                 if(dp[j-1][k-nod[i].y] == -1) continue;
43                 dp[j][k] = max(dp[j][k], dp[j-1][k-nod[i].y]+nod[i].x);
44             }
45         }
46     }
47     int ans = 0;
48     for(int i = 1; i <= MAXY; i ++) {
49         ans = max(ans, min(i,dp[k][i]));
50     }
51     printf("%d\n",ans);
52     return 0;
53 }