PROBLEM LINK:

Contest
Practice

Author:Sergey Kulik
Testers:Vasya Antoniuk and Kevin Atienza
Translators:Vasya Antoniuk (Russian), Team VNOI (Vietnamese) and Hu Zecong (Mandarin)
Editorialist:Kevin Atienza

DIFFICULTY:

Simple

PREREQUISITES:

Loops

PROBLEM:

$N$ students will take $E$ exams. All the students with score strictly greater than at least $(N-K)$ students' total score pass. Each exam's score is an integer between $0$ and $M$, inclusive.

They have already taken $E-1$ exams. Sergey is the $N$th person. Given that he knows the results of the other $N-1$ students in the remaining exam, what is the minimum score he must have to pass? (or determine if it is impossible to pass.)

QUICK EXPLANATION:

• Find all total scores of the first $N-1$ students, and sort them in increasing order. Let the $(N-K)$'th smallest total be $x$.
• Let $y$ be the total score Sergey in the first $E-1$ exams.
• Then Sergey needs at least $\max(0, x - y + 1)$ points. If this exceeds $M$, then the answer is Impossible.

EXPLANATION:

Sergey passes if and only if his total score is strictly greater than the $(N-K)$'th smallest total score. So the answer consists of two parts:

• Computing the $(N-K)$'th smallest total score, and
• Finding the smallest number of points for Sergey to exceed this total score.

The first part can easily be done by first computing the $N-1$ total scores of the other students using a nested loop, and then sorting. The $(N-K)$'th element of the sorted array is what we want! This runs in $O(N \log N + NE)$ time, but we also note that this can be reduced to just $O(NE)$ by using a linear-time selection algorithm.

For the second part, suppose the $(N-K)$'th smallest total is $x$, and Sergey's total score in the first $E-1$ exams is $y$. ($y$ can be computed with a simple loop). Then we must find the minimum score which, when added to $y$, becomes strictly greater than $x$. If $y > x$ already, then this is $0$. Otherwise (i.e., if $y\le x$), then we need to find an integer $T$ such that $y + T > x$. Clearly this is minimized if the sum is as small as possible, but the smallest integer $> x$ is $x + 1$, so $y + T$ must be equal to $x + 1$. Thus, $T = x - y + 1$. Finally, note that an exam score is at most $M$, so we must check if $T \le M$, because if $T > M$, then we conclude that Sergey will not pass.

Here's an implementation of this algorithm in C++:

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

long long totals[10011];

// select the I'th smallest element from totals[0]...totals[N-1]
long long select(int N, int I) {
    sort(totals, totals+N);
    return totals[I-1];
}

int main() {
    int T;
    cin >> T;
    for (int cas = 1; cas <= T; cas++) {
        int N, K, E, M;
        cin >> N >> K >> E >> M;
        for (int i = 0; i < N-1; i++) {
            long long total = 0;
            for (int j = 0; j < E; j++) {
                long long score;
                cin >> score;
                total += score;
            }
            totals[i] = total;
        }
        long long x = select(N-1, N-K);
        long long y = 0;
        for (int j = 0; j < E-1; j++) {
            long long score;
            cin >> score;
            y += score;
        }
        long long answer = max(0LL, x - y + 1);
        if (answer > M) {
            cout << "Impossible" << endl;
        } else {
            cout << answer << endl;
        }
    }
}

Note that we don't store the individual exam scores. Rather we compute the total of each student on the fly, and just store the totals.

Time Complexity:

$O(N \log N + NE)$

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