Given an integer N, the task is to find the total number of ways a sequence can be formed consisting of distinct consecutive odd integers that add up to N.
Examples:
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
Approach: The idea to solve the problem is based on the idea of sum of first K consecutive odd integers:
- 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 xth 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.
Follow the steps mentioned below to implement the above observation:
- Iterate through all pairs of divisors, such that their product is N.
- If the sum of such a pair of divisors is even, increment the count of answer by 1.
- Return the final count at the end.
Below is the implementation of the above approach.
// C++ program for above approach: #include <bits/stdc++.h> 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 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. |
# 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# 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. |
<script> // 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 </script> |
3
Time Complexity: O(√N)
Auxiliary Space: O(1)