Codeforces-837D Round Subset（dp）

D. Round Subset

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., an (1 ≤ ai ≤ 1018).

Output

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

Examples

input

3 250 4 20

output

3

input

5 315 16 3 25 9

output

3

input

3 39 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.

这道题莫名其妙可以用n^2*5000的复杂度过（果然CF很神奇）

dp[t][i][j]表示将第i个数作为第t个取出来，现在有j个5，dp值代表有多少个2

因为开不了这么大的数组，第二维用滚动数组省去，最后答案就是max(min(dp[k][j],j))

#include<stdio.h>#include<iostream>#include<algorithm>#include<stdlib.h>#include<math.h>#include<string.h>#include<set>#include<vector>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define first x#define second y#define eps 1e-5using namespace std;typedef long long LL;typedef pair<int, int> PII;const int inf = 0x3f3f3f3f;const int MX = 200 + 5;struct node {    int x, y;    bool operator<(const node& _A)const {        if (x != _A.x) return x < _A.x;        return y < _A.y;    }} a[MX];int dp[MX][MX * 30];int main() {    //freopen("in.txt", "r", stdin);    int n, k;    scanf("%d%d", &n, &k);    int tot = 0;    for (int i = 1; i <= n; i++) {        LL t;        scanf("%I64d", &t);        int cnt1 = 0, cnt2 = 0;        while (t % 5 == 0) t /= 5, cnt1++;        while (t % 2 == 0) t /= 2, cnt2++;        a[i].x = cnt1;        a[i].y = cnt2;        tot += a[i].x;    }    memset(dp, -1, sizeof(dp));    for (int i = 0; i <= n; i++) dp[i][0] = 0;    for (int i = 1; i <= n; i++) {        for (int t = min(i, k); t > 0; t--) {            for (int j = tot; j >= a[i].x; j--) {                if (dp[t - 1][j - a[i].x] >= 0) dp[t][j] = max(dp[t][j], dp[t - 1][j - a[i].x] + a[i].y);            }        }    }    int ans = 0;    for (int i = 1; i <= tot; i++) ans = max(ans, min(dp[k][i], i));    printf("%d\n", ans);    return 0;}