生成总和为 N 的不同连续奇数序列的方法计数

给定一个整数N ，任务是找出由加起来为N的不同连续奇数整数组成的序列的总数。

例子：

Input: N = 45
Output: 3
Explanation: 3 ways to choose distinct consecutive odd numbers that add up to 45 are – 
{5, 7, 9, 11, 13}, {13, 15, 17} and {45}.

Input : N = 20
Output : 1
Explanation: 9 and 11 are the only consecutive odd numbers whose sum is 20

方法：解决问题的思想是基于前K个连续奇数之和的思想：

• The sum of first K consecutive odd integers is K².
• Let there be a sequence of consecutive odd integers from (y+1)th odd number to x^th odd number (x > y), whose sum is N.
• Then, x² – y² = N or (x + y) * (x – y) = N.
• Let a and b be two divisors of N. Therefore, a * b=N.
  • Hence, x + y = a & x – y = b
  • Solving these two, we get x = (a + b) / 2.
• This implies, if (a + b) is even, then x and y would be integral, which means there would exist a sequence of consecutive odd integers that adds up to N.

按照下面提到的步骤来实现上述观察：

• 遍历所有除数对，使其乘积为N 。
• 如果这样一对除数之和是偶数，则将答案的计数加 1。
• 最后返回最终计数。

下面是上述方法的实现。

C++

// C++ program for above approach:
#include 
using namespace std;
 
// Function to calculate
// Number of sequence of odd integers that
// Contains distinct consecutive odd integers
// That add up to N.
int numberOfSequences(int N)
{
    // Initializing count variable by 0,
    // That stores the number of sequences
    int count = 0;
 
    // Iterating through all divisors of N
    for (int i = 1; i * i <= N; i++) {
        if (N % i == 0) {
 
            // If sum of the two divisors
            // Is even, we increment
            // The count by 1
            int divisor1 = i;
            int divisor2 = N / i;
            int sum = divisor1 + divisor2;
            if (sum % 2 == 0) {
                count++;
            }
        }
    }
 
    // Returning total count
    // After completing the iteration
    return count;
}
 
// Driver Code
int main()
{
    int N = 45;
 
    // Function call
    int number_of_sequences = numberOfSequences(N);
    cout << number_of_sequences;
    return 0;
}

Java

// JAVA program to check whether sum
// Is equal to target value
// After K operations
import java.util.*;
class GFG
{
 
  // Function to calculate
  // Number of sequence of odd integers that
  // Contains distinct consecutive odd integers
  // That add up to N.
  static int numberOfSequences(int N)
  {
    // Initializing count variable by 0,
    // That stores the number of sequences
    int count = 0;
 
    // Iterating through all divisors of N
    for (int i = 1; i * i <= N; i++) {
      if (N % i == 0) {
 
        // If sum of the two divisors
        // Is even, we increment
        // The count by 1
        int divisor1 = i;
        int divisor2 = N / i;
        int sum = divisor1 + divisor2;
        if (sum % 2 == 0) {
          count++;
        }
      }
    }
 
    // Returning total count
    // After completing the iteration
    return count;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 45;
 
    // Function call
    int number_of_sequences = numberOfSequences(N);
    System.out.print(number_of_sequences);
  }
}
 
// This code is contributed by sanjoy_62.

Python3

# Python code for the above approach
import math
 
# Function to calculate
# Number of sequence of odd integers that
# Contains distinct consecutive odd integers
# That add up to N.
def numberOfSequences(N):
 
    # Initializing count variable by 0,
    # That stores the number of sequences
    count = 0;
 
    # Iterating through all divisors of N
    for i in range(1,math.ceil(math.sqrt(N))):
        if (N % i == 0):
           
            # If sum of the two divisors
            # Is even, we increment
            # The count by 1
            divisor1 = i;
            divisor2 = N //i;
            sum = divisor1 + divisor2;
            if (sum % 2 == 0):
                count = count + 1
             
    # Returning total count
    # After completing the iteration
    return count;
 
# Driver Code
N = 45;
 
# Function call
number_of_sequences = numberOfSequences(N);
print(number_of_sequences);
    
# This code is contributed by Potta Lokesh

C#

// C# program for above approach:
using System;
class GFG {
 
  // Function to calculate
  // Number of sequence of odd integers that
  // Contains distinct consecutive odd integers
  // That add up to N.
  static int numberOfSequences(int N)
  {
    // Initializing count variable by 0,
    // That stores the number of sequences
    int count = 0;
 
    // Iterating through all divisors of N
    for (int i = 1; i * i <= N; i++) {
      if (N % i == 0) {
 
        // If sum of the two divisors
        // Is even, we increment
        // The count by 1
        int divisor1 = i;
        int divisor2 = N / i;
        int sum = divisor1 + divisor2;
        if (sum % 2 == 0) {
          count++;
        }
      }
    }
 
    // Returning total count
    // After completing the iteration
    return count;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 45;
 
    // Function call
    int number_of_sequences = numberOfSequences(N);
    Console.Write(number_of_sequences);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program for above approach:
 
    // Function to calculate
    // Number of sequence of odd integers that
    // Contains distinct consecutive odd integers
    // That add up to N.
    const numberOfSequences = (N) => {
     
        // Initializing count variable by 0,
        // That stores the number of sequences
        let count = 0;
 
        // Iterating through all divisors of N
        for (let i = 1; i * i <= N; i++) {
            if (N % i == 0) {
 
                // If sum of the two divisors
                // Is even, we increment
                // The count by 1
                let divisor1 = i;
                let divisor2 = parseInt(N / i);
                let sum = divisor1 + divisor2;
                if (sum % 2 == 0) {
                    count++;
                }
            }
        }
 
        // Returning total count
        // After completing the iteration
        return count;
    }
 
    // Driver Code
    let N = 45;
 
    // Function call
    let number_of_sequences = numberOfSequences(N);
    document.write(number_of_sequences);
 
// This code is contributed by rakeshsahni
 

输出

3

时间复杂度： O(√N)
辅助空间： O(1)