📘 Day 10 – Teacher's Day Challenge | Next Round Qualification

🗓️ Date: August 03, 2025
🧩 Platform: Codeforces
🔢 Problems Solved Today: 1
🎯 Focus Topic: Arrays, Conditional Logic, Ranking System

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🔗 Table of Contents

• Problem 1 – Next Round

• Daily Summary

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🧩 Problem 1 – Next Round

📚 Difficulty: Easy
🧠 Concepts: Array Traversal, Indexing, Score Threshold

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📄 Problem Statement (Summary):
You are given the scores of n participants in non-increasing order and an integer k.
A participant advances to the next round if:

• Their score is ≥ score of the k-th participant, and

• Their score is greater than 0.
  Count how many participants will advance.

📝 Example:
Input:

8 5  
10 9 8 7 7 7 5 5

Output:

6

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💡 Approach:

• Input n and k, and store n scores in a vector.

• The score of the k-th participant is nums[k-1].

• Traverse all scores:

  • If current score ≥ nums[k-1] and > 0, count it.

• Print the final count.

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🧪 Code (C++):

#include <iostream>
#include <vector>
using namespace std;
int main()
{
    int n, k, c = 0;
    cin >> n >> k;
    vector<int> nums(n);
    for (int i = 0; i < nums.size(); i++)
        cin >> nums[i];
    for (int i = 0; i < n; i++) {
        if (nums[i] >= nums[k - 1] && nums[i] > 0)
            c++;
    }
    cout << c;
    return 0;
}

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📸 Submission Screenshot:
(Attach your accepted screenshot here using ➕ image in editor)

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✅ Key Takeaways:

• Carefully access the k-th element using k-1 index (0-based indexing).

• Ensure the score is positive before counting.

• The non-increasing order guarantees we can compare directly from the front.

• Clean array traversal with simple conditions is key.

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📈 Daily Summary

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🏷️ Tags:
#codeforces #arrays #ranking #contestlogic #cpp #dsa #43DaysChallenge