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A. Kill Two Birds with One Stone
verdict.run/gym/106259/problem/A
A.Kill Two Birds with One Stone

Problem Statement

Why are lots of things in twos? Hands on clocks, and gloves, and shoes, Scissor-blades, and water taps, Collar studs, and luggage straps...

You are given a binary matrix M of size n × m, where rows are numbered from 1 to n from top to bottom and columns are numbered from 1 to m from left to right.

Your task is to determine whether it is possible to transform the matrix into a null matrix (i.e., a matrix where all entries are 0).

INPUT

The first line contains t (1 ≤ t ≤ 104) — the number of test cases.

OUTPUT

For each test case, output YES if possible, NO otherwise.

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GNU C++20 (64)
Run TestsSubmit
1#include <bits/stdc++.h>
2using namespace std;
3 
4void solve() {
5 int n; cin >> n;
6 vector<int> a(n);
7 for (auto& x : a) cin >> x;
8 sort(a.begin(), a.end());
9 cout << a[0] << '\n';
10}
11 
12int main() {
13 int t; cin >> t;
14 while (t--) solve();
15}
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How do I solve this problem? I'm stuck on the greedy approach.

Great question! Let's think about this step by step.

Hint 1: Consider sorting the array first. What property does the sorted array give you?

Key insight: After sorting, the minimum difference between any two elements must be between adjacent elements.

C++@ lines 5-8
sort(a.begin(), a.end());
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Testcase
Test ResultCodeforces
Case 1Case 2Case 3
Input
3
5 2 8 1 4
3 3 7
Expected Output
1
3
3
All test cases passed 46ms

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V
verdict.run
Compute ceiling division
solution.cpp
1#include <bits/stdc++.h>
2using namespace std;
3int main() {
4 long long n, m, a;
5 cin >> n >> m >> a;
6 long long rows = (n+a-1)/a;
7 long long cols = (m+a-1)/a;
8 cout << rows*cols << endl;
9 return 0;
10}

Here is the core logic. We compute the number of rows of tiles as n plus a minus 1, integer divided by a...

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