Thief in a Shop+dp计数

Thief in a Shop
time limit per test
5 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output

A thief made his way to a shop.

As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product of kind i is ai.

The thief is greedy, so he will take exactly k products (it's possible for some kinds to take several products of that kind).

Find all the possible total costs of products the thief can nick into his knapsack.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 1000) — the number of kinds of products and the number of products the thief will take.

The second line contains n integers ai (1 ≤ ai ≤ 1000) — the costs of products for kinds from 1 to n.

Output

Print the only line with all the possible total costs of stolen products, separated by a space. The numbers should be printed in the ascending order.

Examples
Input
Copy
3 2
1 2 3
Output
Copy
2 3 4 5 6
Input
Copy
5 5
1 1 1 1 1
Output
Copy
5
Input
Copy
3 3
3 5 11
Output
Copy
9 11 13 15 17 19 21 25 27 33


#include <iostream>
#include <fstream>
#include <set>
#include <map>
#include <string>
#include <vector>
#include <bitset>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cassert>
#include <queue>


typedef long long ll;
typedef long double ld;

using namespace std;

int n, k;
int dp[1000010];
vector<int> vv;

int main() {
    freopen("in.txt","r",stdin);
	cin >> n >> k;
	int mn = 10000;
	for (int i = 0; i < n; ++i) {
		int a;
		cin >> a;
		mn = min(mn, a);
		vv.push_back(a);
	}
	for (int i = 0; i < (int)vv.size(); ++i)
		vv[i] -= mn;
	sort(vv.begin(), vv.end());
	vv.resize(unique(vv.begin(), vv.end()) - vv.begin());
	int mx = k * vv.back();
	for (int j = 0; j <= mx; ++j)
		dp[j] = k + 1;
	dp[0] = 0;
	for (int i = 0; i < (int)vv.size(); ++i) {
		int x = vv[i];
		if (x == 0)
			continue;
		for (int j = 0; j <= mx - x; ++j)
			dp[j + x] = min(dp[j + x], dp[j] + 1);
	}
	for (int i = 0; i <= mx; ++i)
		if (dp[i] <= k)
			printf("%d ", mn * k + i);
	return 0;
}

猜你喜欢

转载自blog.csdn.net/ujn20161222/article/details/80507599