Given an integer N, the task is to count all arithmetic progression with common difference equal to 1 and the sum of all its elements equal to N.
Examples:
Input: N = 12
Output: 3
Explanation:
Following three APs satisfy the required conditions:
- {3, 4, 5}
- {−2, −1, 0, 1, 2, 3, 4, 5}
- {−11, −10, −9, …, 10, 11, 12]}
Input: N = 963761198400
Output: 1919
Approach: The given problem can be solved using the following properties of AP:
- The sum of an AP series having N terms with first term as A and common difference 1 is given by the formula S = (N / 2 )* ( 2 *A + N − 1).
- Solving the above equation:
S = N / 2 [2 *A + N − 1].
Rearranging this and multiplying both sides by 2
=> 2 * S = N * (N + 2 * A − 1)Rearranging further
=> A = ((2 * N / i ) − i + 1) / 2.Now, iterate over all the factors of 2 * N, and check for each factor i, whether (2 * N/i) − i + 1 is even or not.
Follow the steps below to solve the problem:
- Initialize a variable, say count, to store the number of AP series possible having given conditions.
- Iterate through all the factors of 2 * N and for every ith factor, check if (2 * N / i) − i + 1 is even.
- If found to be true, increment the count.
- Print count – 1 as the required answer, as the sequence consisting only of N needs to be discarded.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count all possible// AP series with common difference// 1 and sum of elements equal to Nvoid countAPs(long long int N){ // Stores the count of AP series long long int count = 0; // Traverse through all factors of 2 * N for (long long int i = 1; i * i <= 2 * N; i++) { long long int res = 2 * N; if (res % i == 0) { // Check for the given conditions long long int op = res / i - i + 1; if (op % 2 == 0) { // Increment count count++; } if (i * i != res and (i - res / i + 1) % 2 == 0) { count++; } } } // Print count - 1 cout << count - 1 << "\n";}// Driver Codeint main(){ // Given value of N long long int N = 963761198400; // Function call to count // required number of AP series countAPs(N);} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Function to count all possible // AP series with common difference // 1 and sum of elements equal to N static void countAPs(long N) { // Stores the count of AP series long count = 0; // Traverse through all factors of 2 * N for (long i = 1; i * i <= 2 * N; i++) { long res = 2 * N; if (res % i == 0) { // Check for the given conditions long op = res / i - i + 1; if (op % 2 == 0) { // Increment count count++; } if (i * i != res && (i - res / i + 1) % 2 == 0) { count++; } } } // Print count - 1 System.out.println(count - 1); } // Driver code public static void main(String[] args) { // Given value of N long N = 963761198400L; // Function call to count // required number of AP series countAPs(N); }}// This code is contributed by Kingash. |
Python3
# Python3 program to implement# the above approach# Function to count all possible# AP series with common difference# 1 and sum of elements equal to Ndef countAPs(N) : # Stores the count of AP series count = 0 # Traverse through all factors of 2 * N i = 1 while(i * i <= 2 * N) : res = 2 * N if (res % i == 0) : # Check for the given conditions op = res / i - i + 1 if (op % 2 == 0) : # Increment count count += 1 if (i * i != res and (i - res / i + 1) % 2 == 0) : count += 1 i += 1 # Print count - 1 print(count - 1)# Driver Code# Given value of NN = 963761198400# Function call to count# required number of AP seriescountAPs(N)# This code is contributed by sanjoy_62. |
C#
// C# program for above approachusing System;public class GFG { // Function to count all possible // AP series with common difference // 1 and sum of elements equal to N static void countAPs(long N) { // Stores the count of AP series long count = 0; // Traverse through all factors of 2 * N for (long i = 1; i * i <= 2 * N; i++) { long res = 2 * N; if (res % i == 0) { // Check for the given conditions long op = res / i - i + 1; if (op % 2 == 0) { // Increment count count++; } if (i * i != res && (i - res / i + 1) % 2 == 0) { count++; } } } // Print count - 1 Console.WriteLine(count - 1); } // Driver code public static void Main(String[] args) { // Given value of N long N = 963761198400L; // Function call to count // required number of AP series countAPs(N); }}// This code is contributed by sanjoy_62. |
Javascript
<script>// Javascript program for the above approach // Function to count all possible // AP series with common difference // 1 and sum of elements equal to N function countAPs(N) { // Stores the count of AP series let count = 0; // Traverse through all factors of 2 * N for (let i = 1; i * i <= 2 * N; i++) { let res = 2 * N; if (res % i == 0) { // Check for the given conditions let op = res / i - i + 1; if (op % 2 == 0) { // Increment count count++; } if (i * i != res && (i - res / i + 1) % 2 == 0) { count++; } } } // Print count - 1 document.write(count - 1); }// Driver code // Given value of N let N = 963761198400; // Function call to count // required number of AP series countAPs(N); </script> |
Output:
1919
Time Complexity: O(√N)
Auxiliary Space: O(1)
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