# USACO 2020 January Contest, Silver Problem 3. Wormhole Sort

### USACO 2020 January Contest, Silver Problem 3. Wormhole Sort

Bessie and her little sister Elsie are picking berries in Farmer John's berry patch. Farmer John's patch has exactly NN berry trees (1N10001≤N≤1000); tree ii contains exactly BiBi berries (1Bi10001≤Bi≤1000). Bessie has exactly KK baskets (1K10001≤K≤1000KK even). Each basket can hold as many berries from a single tree as Bessie wants, but cannot contain berries from two different trees as their flavors will clash with each other. Baskets may remain empty.

Bessie wants to maximize the number of berries she collects. However, Farmer John wants Bessie to share with her little sister, and so Bessie will have to give Elsie the K/2K/2 baskets with the largest number of berries. This means that Elsie may even end up with more berries than Bessie, which is very unfair, but unfortunately, sibling dynamics are not always fair.

Help Bessie figure out the maximum number of berries she can collect.

#### SCORING:

• Test cases 1-4 satisfy K10.K≤10.
• Test cases 5-11 satisfy no additional constraints.

#### INPUT FORMAT (file berries.in):

The first line of input contains space-separated integers NN and KK.The second line contains NN space-separated integers B1,B2,,BN.B1,B2,…,BN.

#### OUTPUT FORMAT (file berries.out):

A single line with the answer.

#### SAMPLE INPUT:

5 4
3 6 8 4 2


#### SAMPLE OUTPUT:

8


If Bessie fills

• one basket with 6 berries from tree 2
• two baskets, each with 4 berries from tree 3
• one basket with 4 berries from tree 4

then she receives two baskets each with 4 berries, giving her 8 berries in total.

Problem credits: Nathan Pinsker

<h3> USACO 2020 January Contest, Silver Problem 3. Wormhole Sort 题解(翰林国际教育提供，仅供参考)</h3>
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(Analysis by Dhruv Rohatgi, Benjamin Qi)

Small KK:

After sorting the trees in decreasing order of BiBi, we don't need to consider trees outside of the first K.K. Furthermore, if we decide to select b>0b>0 baskets from tree i,i, then each basket should have either Bib⌊Bib⌋ or Bib+1⌊Bib⌋+1 berries. Using these observations, we can do some sort of backtracking.

Full Solution:

Let bb the minimum number of berries in one of the buckets that Elsie receives. Without loss of generality, we can assume that all of Elsie's buckets contain exactly bb berries. Now our goal is to maximize the number of berries placed into KK buckets of size at most bb such that at least K2K2 buckets have exactly bb berries inside.

Consider a single tree's allotment into the buckets in an optimal solution. There's no point having multiple buckets with less than bb berries from this tree. So all buckets will have exactly bb berries aside from at most one.

Thus, it's clearly optimal to repeatedly fill buckets of size exactly bb until we run out of buckets or all trees have less than bb berries remaining. If we still have buckets to fill, sort the remaining trees by Bi(modb)Bi(modb) and iterate from the largest to the smallest value.

We can repeat this procedure for each b=0max(Bi),b=0…max(Bi), which runs in O(max(Bi)NlogN)O(max(Bi)⋅Nlog⁡N) time.

Dhruv Rohatgi's code:

#include <iostream>
#include <algorithm>
using namespace std;

int N,K;
int A[100000];
int mod;

bool cmp(int a,int b)
{
return (a%mod) > (b%mod);
}

int main()
{
freopen("berries.in","r",stdin);
freopen("berries.out","w",stdout);
cin >> N >> K;
int M = 0;
for(int i=0;i<N;i++)
{
cin >> A[i];
M = max(M, A[i]);
}
int best = 0;
for(int b=1;b <= M;b++)
{
int full = 0;
for(int i=0;i<N;i++)
full += A[i]/b;
if(full < K/2)
break;
if(full >= K)
{
best = max(best, b*(K/2));
continue;
}
mod = b;
sort(A, A+N, cmp);
int cur = b*(full - K/2);
for(int i=0;i<N && i+full<K;i++)
cur += A[i]%b;
best = max(best,cur);
}
cout << best << '\n';
}

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