C++ Coding Test Course, Finding the Sum of Consecutive Natural Numbers

In this lecture, we will address the problem of finding the sum of consecutive natural numbers. This problem is a good one for evaluating both algorithmic and mathematical thinking skills. For example, we will examine the problem of finding the number of ways to express a given integer N as a sum of consecutive natural numbers.

Problem Description

Given an integer N, write a function to find the number of ways N can be expressed as a sum of consecutive natural numbers.

• Input: A single integer N (1 ≤ N ≤ 10⁶)
• Output: The number of ways to express N as a sum of consecutive natural numbers

Example Problem

Input: 15
Output: 4
Explanation: 15 can be expressed as a sum of consecutive natural numbers as follows:
- 15 = 15
- 15 = 7 + 8
- 15 = 4 + 5 + 6
- 15 = 1 + 2 + 3 + 4 + 5

Approach to the Problem

To solve this problem, it is important to understand the mathematical properties of the sum of consecutive natural numbers. The sum S of consecutive natural numbers can be expressed as follows:

S = a + (a + 1) + (a + 2) + ... + (a + (k-1)) = k * a + (0 + 1 + 2 + ... + (k-1))
S = k * a + (k * (k - 1)) / 2

Here, a is the first natural number, and k is the number of consecutive natural numbers. By rearranging the above equation, we can derive the following relationship:

N = k * a + (k * (k - 1)) / 2
N - (k * (k - 1)) / 2 = k * a

Therefore, if N – (k * (k – 1)) / 2 is a multiple of k, we can find the natural number a. If this condition is met, we can express N as a sum of k consecutive natural numbers.

Implementation Strategy

The specific algorithm to solve the problem is as follows:

1. Start with k as 1 and progress until it exceeds N.
2. For each k, calculate N – (k * (k – 1)) / 2.
3. Check if the calculated value is a multiple of k.
4. If it is a multiple, increment the count of cases.

C++ Code Implementation

Now, let’s implement the above algorithm in C++ code.

#include <iostream>

using namespace std;

int countConsecutiveSum(int N) {
    int count = 0;

    for (int k = 1; k * (k - 1) / 2 < N; ++k) {
        int tmp = N - (k * (k - 1) / 2);
        if (tmp % k == 0) {
            count++;
        }
    }

    return count;
}

int main() {
    int N;
    cout << "Enter N: ";
    cin >> N;

    int result = countConsecutiveSum(N);
    cout << "Number of ways to express as a sum of consecutive natural numbers: " << result << endl;

    return 0;
}

Example Execution

Enter N: 15
Number of ways to express as a sum of consecutive natural numbers: 4

Time Complexity Analysis

The time complexity of this algorithm is O(√N). This is because the maximum value of k is less than N, allowing the loop for k to terminate sooner than it would for N. This level of time complexity is efficient enough for the given input range of the problem.

Conclusion

In this lecture, we explored the problem of finding the sum of consecutive natural numbers, learning the basic concepts of algorithm design and C++ implementation. I hope that by engaging with problems like this, you will continue to build knowledge that can be applied in actual coding tests.