**Problem Statement**

There are k friends who want to buy N flowers. Each flower has some cost ci. But there is condition that the price of flowers changes for customer who had bought flowers before. More precisely if a customer has already bought x flowers, he should pay (x+1)*ci dollars to buy flower number i. What is the minimum amount you have to pay to buy all N flowers?

Ex Input : Output :

3 3 13

2 5 6

So first line contains n and k and next line contain ci, cost of ith flower.

Algorithm

1) Arrange all the cost in an increasing sequence…d1,d2..dn

2) Allot from end nth flower to friend 1, n-1 flower to friend 2 and so on..until k after which cost factor would be 2 and hence flower n-k would cost 2*d(n-k) to friend 1 and n-k-1 flower will cost 2*d(n-k-1) to friend 2 and so on until all flowers are bought by somebody.

If you have a different approach in mind, please share it in comments section.

Solution #include #include using namespace std; int main() { int n,k; cin >> n >> k; if(k > n) k=n; int i,x; multiset s; multiset::iterator it; for(i=0;i<n;i++) { cin >> x; s.insert(x); } //start calculating cost int factor = 0; unsigned long long int cost=0; int count = 0; it = s.end(); it--; while(count!=n) { if(count%k==0) factor+=1; cost+=(*it * factor); it--; count++; } cout << cost << endl; cin >> cost; }

Time Complexity

O(nlogn) to sort n cost numbers and O(n) to calculate the cost.

O(n+nlogn) => O(nlogn)

Hence time complexity will be O(nlogn)

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