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.
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.
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.
3 2
1 2 3
2 3 4 5 6
5 5
1 1 1 1 1
5
3 3
3 5 11
9 11 13 15 17 19 21 25 27 33
题目大意：给你n个数，让你选k次，可重复选，输出所有可能值。
思路：如果简单的用个二维dp[i][j],表示选i次价值为j是否存在，但这样显然会爆，10^12，那么改良下，用do[i]表示价值为i时，最少选几次，剩余的次数可以让其中的最小值去补。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[1000006],s[1003];
const int INF=0x3f3f3f3f;
const int MAX=999999;
int main()
{ int n,k;
while(cin>>n>>k)
{
for(int i=0;i<n;i++)
scanf("%d",&s[i]);
sort(s,s+n);
int t=s[0];
for(int i=0;i<n;i++)
s[i]-=t;
for(int i=0;i<=k*s[n-1];i++)
dp[i]=k+1;
dp[0]=0;
for(int i=0;i<=k*s[n-1];i++)
{
for(int j=0;j<n;j++)
if(i>=s[j])
dp[i]=min(dp[i-s[j]]+1,dp[i]);
}
for(int i=0;i<=k*s[n-1];i++)
if(dp[i]<=k)
cout<<i+k*t<<" ";
cout<<endl;
}
}