PROBLEM LINK

Practice
Contest

Panel Members

Problem Setter:Suhash
Problem Tester:
Editorialist:Sunny Aggarwal
Contest Admin:
Russian Translator:
Mandarin Translator:
Vietnamese Translator:
Language Verifier:

DIFFICULTY:

Easy

PREREQUISITES:

Basic Mathematics, Prefix Sum, Sorting, Dynamic Programming.

PROBLEM:

Given a list of $N$ coins of possibly different denominations. We can pay amount equivalent to any $1$ coin and can acquire that coin. In addition once we have paid for a coin, we can choose atmost $K$ more coins and can acquire those for free. What is the minimum amount required to acquire all the $N$ coins for a given $K$ ?

Note that a coin can be acquired only once.

EXPLANATION

It is easy to notice that at a cost of $1$ coin, we can acquire at most $K+1$ coins. Therefore, in order to acquire all the $N$ coins we will be choosing $\lceil N/(K+1)\rceil$ coins and the cost of choosing coins will be minimum if we choose smallest $\lceil N/(K+1)\rceil$ coins. Smallest $\lceil N/(K+1)\rceil$ coins can be found by simply sorting all the $N$ values in increasing order.

C++ code:

int main() {
    int n, k;
    cin >> n >> k;
    int arr[n];
    for(int i=0; i<=n-1; i++) {
        cin >> arr[i];
    }
    sort(arr, arr+n);
    int coins_needed = ceil(1.0*n/(k+1));
    int ans = 0;
    for(int i=0; i<=coins_needed-1; i++) {
        ans += arr[i];
    }
    cout << ans << "\n";
    return 0;
}

As we are asked to find the above answer for many different values of $K$, we have to compute it fast. For the purpose to serve, we can maintain a prefix sum array after sorting all the $N$ values and can answer queries easily.

C++ code:

int main() {
    int n, q, k;
    cin >> n >> q;
    int arr[n];
    for(int i=0; i<=n-1; i++) {
        cin >> arr[i];
    }
    sort(arr, arr+n);
        for(int i=1; i<=n-1; i++) {
        arr[i] += arr[i-1];
    }
    while( q-- ) {
        cin >> k;
        int coins_needed = ceil(1.0*n/(k+1));
        int ans = arr[coins_needed-1];
        cout << ans << "\n";
    }
    return 0;
}

COMPLEXITY

$O(Nlog(N) + Q)$

SIMILIAR PROBLEMS

Ruspa And Game

Query Over Matrix

Minimum Maximum