Dado un número s (1 <= s <= 1000000000). Si este número es la suma de los primeros n números naturales, imprima, de lo contrario, imprima -1.
Ejemplos:
Input: s = 10 Output: n = 4 Explanation: 1 + 2 + 3 + 4 = 10 Input: s = 17 Output: n = -1 Explanation: 17 can't be expressed as a sum of consecutive from 1.
Método 1 (Simple):
C++
// C++ program for above implementation
#include <iostream>
using namespace std;
// Function to find no. of elements
// to be added from 1 to get n
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
int main()
{
int s = 15;
int n = findS(s);
n == -1 ? cout << "-1"
: cout << n;
return 0;
}
Java
// Java program for above implementation
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(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 n
def 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 code
s = 15
n = findS (s)
if n == -1:
print("-1")
else:
print(n)
# This code is contributed by "Abhishek Sharma 44".
C#
// C# program for above implementation
using 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 n
function 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";
else
echo $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>
C++
// C++ program for finding s such
// that sum from 1 to s equals to n
#include <iostream>
using namespace std;
// Function to find no. of elements
// to be added to get s
int findS(int s)
{
int l = 1, r = (s / 2) + 1;
// Apply Binary search
while (l <= r) {
// Find mid
int mid = (l + r) / 2;
// find sum of 1 to mid natural numbers
// using formula
int sum = mid * (mid + 1) / 2;
// If sum is equal to n
// return mid
if (sum == s)
return mid;
// If greater than n
// do r = mid-1
else if (sum > s)
r = mid - 1;
// else do l = mid + 1
else
l = mid + 1;
}
// If not possible, return -1
return -1;
}
// Drivers code
int main()
{
int s = 15;
int n = findS(s);
n == -1 ? cout << "-1"
: cout << n;
return 0;
}
Java
// java program for finding s such
// that sum from 1 to s equals to n
import java.io.*;
public class GFG {
// Function to find no. of elements
// to be added to get s
static int findS(int s)
{
int l = 1, r = (s / 2) + 1;
// Apply Binary search
while (l <= r) {
// Find mid
int mid = (l + r) / 2;
// find sum of 1 to mid natural
// numbers using formula
int sum = mid * (mid + 1) / 2;
// If sum is equal to n
// return mid
if (sum == s)
return mid;
// If greater than n
// do r = mid-1
else if (sum > s)
r = mid - 1;
// else do l = mid + 1
else
l = mid + 1;
}
// If not possible, return -1
return -1;
}
// Drivers code
static public 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 vt_m.
Python3
# python program for finding s such # that sum from 1 to s equals to n # Function to find no. of elements # to be added to get s def findS(s): l = 1 r = int(s / 2) + 1 # Apply Binary search while (l <= r) : # Find mid mid = int((l + r) / 2) # find sum of 1 to mid natural numbers # using formula sum = int(mid * (mid + 1) / 2) # If sum is equal to n # return mid if (sum == s): return mid # If greater than n # do r = mid-1 elif (sum > s): r = mid - 1 # else do l = mid + 1 else: l = mid + 1 # If not possible, return -1 return -1 s = 15 n = findS(s) if(n == -1): print( "-1") else: print( n ) # This code is contributed by Sam007
C#
// C# program for finding s such
// that sum from 1 to s equals to n
using System;
public class GFG {
// Function to find no. of elements
// to be added to get s
static int findS(int s)
{
int l = 1, r = (s / 2) + 1;
// Apply Binary search
while (l <= r) {
// Find mid
int mid = (l + r) / 2;
// find sum of 1 to mid natural
// numbers using formula
int sum = mid * (mid + 1) / 2;
// If sum is equal to n
// return mid
if (sum == s)
return mid;
// If greater than n
// do r = mid-1
else if (sum > s)
r = mid - 1;
// else do l = mid + 1
else
l = mid + 1;
}
// If not possible, return -1
return -1;
}
// Drivers code
static public 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 finding
// s such that sum from 1
// to s equals to n
// Function to find no.
// of elements to be added
// to get s
function findS($s)
{
$l = 1;
$r = 1 + (int)$s / 2;
// Apply Binary search
while ($l <= $r)
{
// Find mid
$mid = (int)(($l + $r) / 2);
// find sum of 1 to mid natural
// numbers using formula
$sum = (int)($mid *
($mid + 1) / 2);
// If sum is equal to n
// return mid
if ($sum == $s)
return $mid;
// If greater than n
// do r = mid-1
else if ($sum > $s)
$r = $mid - 1;
// else do l = mid + 1
else
$l = $mid + 1;
}
// If not possible,
// return -1
return -1;
}
// Drivers code
$s = 15;
$n = findS($s);
if($n == -1 )
echo "-1";
else
echo $n;
// This code is contributed by Sam007
?>
Javascript
<script>
// Javascript program for finding s such
// that sum from 1 to s equals to n public
// Function to find no. of elements
// to be added to get s
function findS(s) {
var l = 1, r = parseInt((s / 2) + 1);
// Apply Binary search
while (l <= r) {
// Find mid
var mid = parseInt( (l + r) / 2);
// find sum of 1 to mid natural
// numbers using formula
var sum = mid * parseInt((mid + 1) / 2);
// If sum is equal to n
// return mid
if (sum == s)
return mid;
// If greater than n
// do r = mid-1
else if (sum > s)
r = mid - 1;
// else do l = mid + 1
else
l = mid + 1;
}
// If not possible, return -1
return -1;
}
// Drivers code
var s = 15;
var n = findS(s);
if (n == -1)
document.write("-1");
else
document.write(n);
// This code contributed by Rajput-Ji
</script>
C++
// C++ program of the above
// approach
#include <bits/stdc++.h>
# define ll long long
using namespace std;
// Function to check if the
// s is the sum of first N
// natural number
ll int isvalid(ll int s)
{
// Solution of Quadratic Equation
float k=(-1+sqrt(1+8*s))/2;
// Condition to check if the
// solution is a integer
if(ceil(k)==floor(k))
return k;
else
return -1;
}
// Driver Code
int main()
{
int s = 15;
// Function Call
cout<< isvalid(s);
return 0;
}
Java
// Java program of the above
// approach
import java.util.*;
class GFG{
// Function to check if the
// s is the sum of first N
// natural number
public static int isvalid(int s)
{
// Solution of Quadratic Equation
double k = (-1.0 + Math.sqrt(1 + 8 * s)) / 2;
// Condition to check if the
// solution is a integer
if (Math.ceil(k) == Math.floor(k))
return (int)k;
else
return -1;
}
// Driver code
public static void main(String[] args)
{
int s = 15;
// Function call
System.out.print(isvalid(s));
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 program of the above # approach import math # Function to check if the # s is the sum of first N # natural number def isvalid(s): # Solution of Quadratic Equation k = (-1 + math.sqrt(1 + 8 * s)) / 2 # Condition to check if the # solution is a integer if (math.ceil(k) == math.floor(k)): return int(k) else: return -1 # Driver Code s = 15 # Function Call print(isvalid(s)) # This code is contributed by vishu2908
C#
// C# program of the above
// approach
using System;
class GFG{
// Function to check if the
// s is the sum of first N
// natural number
public static int isvalid(int s)
{
// Solution of Quadratic Equation
double k = (-1.0 + Math.Sqrt
(1 + 8 * s)) / 2;
// Condition to check if the
// solution is a integer
if (Math.Ceiling(k) ==
Math.Floor(k))
return (int)k;
else
return -1;
}
// Driver code
public static void Main(string[] args)
{
int s = 15;
// Function call
Console.Write(isvalid(s));
}
}
// This code is contributed by Chitranayal
Javascript
<script>
// Javascript program of the above
// approach
// Function to check if the
// s is the sum of first N
// natural number
function isvalid(s)
{
// Solution of Quadratic Equation
let k = (-1.0 + Math.sqrt(1 + 8 * s)) / 2;
// Condition to check if the
// solution is a integer
if (Math.ceil(k) == Math.floor(k))
return k;
else
return -1;
}
// Driver code
let s = 15;
// Function call
document.write(isvalid(s));
</script>
Publicación traducida automáticamente
Artículo escrito por Sahil_Chhabra y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA