Найдите, является ли данное число суммой первых n натуральных чисел
Опубликовано: 18 Января, 2022
Дано число s (1 <= s <= 1000000000). Если это число является суммой первых n натуральных чисел, то выведите, иначе выведите -1.
Примеры:
Ввод: s = 10 Выход: n = 4 Объяснение: 1 + 2 + 3 + 4 = 10 Ввод: s = 17 Выход: n = -1 Объяснение: 17 не может быть выражено как сумма последовательных от 1.
Метод 1 (простой):
Начните складывать числа от i = 1 до n.
- Проверяем, равна ли сумма n, возвращаем i.
- В противном случае, если сумма> n, верните -1.
Below is the implementation of the above approach:
C++
// C++ program for above implementation#include <iostream>using namespace std;// Function to find no. of elements// to be added from 1 to get nint findS(int s){ int sum = 0; // Start adding numbers from 1 for (int n = 1; sum < s; n++) { sum += n; // If sum becomes equal to s // return n if (sum == s) return n; } return -1;}// Drivers codeint main(){ int s = 15; int n = findS(s); n == -1 ? cout << "-1" : cout << n; return 0;} |
Java
// Java program for above implementationclass GFG { // Function to find no. of elements // to be added from 1 to get n static int findS(int s) { int sum = 0; // Start adding numbers from 1 for (int n = 1; sum < s; n++) { sum += n; // If sum becomes equal to s // return n if (sum == s) return n; } return -1; } // Drivers code public static void main(String[]args) { int s = 15; int n = findS(s); if(n == -1) System.out.println("-1"); else System.out.println(n); }}//This code is contributed by Azkia Anam. |
Python3
# Python3 program to check if# given number is sum of first n# natural numbers# Function to find no. of elements# to be added from 1 to get ndef findS (s): _sum = 0 n = 1 # Start adding numbers from 1 while(_sum < s): _sum += n n+=1 n-=1 # If sum becomes equal to s # return n if _sum == s: return n return -1# Driver codes = 15n = findS (s)if n == -1: print("-1")else: print(n)# This code is contributed by "Abhishek Sharma 44". |
C#
// C# program for above implementationusing System;class GFG { // Function to find no. of elements // to be added from 1 to get n static int findS(int s) { int sum = 0; // Start adding numbers from 1 for (int n = 1; sum < s; n++) { sum += n; // If sum becomes equal to s // return n if (sum == s) return n; } return -1; } // Drivers code public static void Main() { int s = 15; int n = findS(s); if(n == -1) Console.WriteLine("-1"); else Console.WriteLine(n); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program for above implementation// Function to find no. of elements// to be added from 1 to get nfunction findS($s){ $sum = 0; // Start adding numbers from 1 for ($n = 1; $sum < $s; $n++) { $sum += $n; // If sum becomes equal // to s return n if ($sum == $s) return $n; } return -1;}// Drivers code$s = 15;$n = findS($s);if($n == -1)echo "-1";elseecho $n;// This code is contributed by Sam007?> |
Javascript
<script>// Javascript program for above implementation // Function to find no. of elements// to be added from 1 to get n function findS(s) { var sum = 0; // Start adding numbers from 1 for (n = 1; sum < s; n++) { sum += n; // If sum becomes equal to s // return n if (sum == s) return n; } return -1; } // Drivers code var s = 15; var n = findS(s); if (n == -1) document.write("-1"); else document.write(n);// This code is contributed by Rajput-Ji</script> |
Output
5