Dado un número N , la tarea es verificar si el número N se puede representar como la diferencia de dos cubos consecutivos o no. En caso afirmativo , imprima esos números; de lo contrario, imprima No.
Ejemplos:
Entrada: N = 19
Salida:
Sí
2 3
Explicación:
3 3 – 2 3 = 19Entrada: N = 10
Salida: No
Enfoque: La observación clave en el problema es que un número puede representarse como la diferencia de dos cubos consecutivos si y solo si:
=> N = (K+1) 3 – K 3
=> N = 3*K 2 + 3*K + 1
=> 12*N = 36*K 2 + 36*K + 12
=> 12*N = ( 6*K + 3) 2 + 3
=> 12*N – 3 = (6*K + 3) 2
lo que significa que (12*N – 3) debe ser un cuadrado perfecto para descomponer N en la diferencia de dos cubos consecutivos.
Por lo tanto, si la condición anterior es cierta, imprimiremos los números usando un ciclo for para verificar para qué valor de i si (i+1) 3 – i 3 = N e imprimiremos el número i y i + 1 .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the two consecutive
// numbers whose difference is N
void print(int N)
{
// Iterate in the range [0, 10^5]
for (int i = 0; i < 100000; i++) {
if (pow(i + 1, 3)
- pow(i, 3)
== N) {
cout << i << ' ' << i + 1;
return;
}
}
}
// Function to check if N is a
// perfect cube
bool isPerfectSquare(long double x)
{
// Find floating point value of
// square root of x.
long double sr = sqrt(x);
// If square root is an integer
return ((sr - floor(sr)) == 0);
}
// Function to check whether a number
// can be represented as difference
// of two consecutive cubes
bool diffCube(int N)
{
// Check if 12 * N - 3 is a
// perfect square or not
return isPerfectSquare(12 * N - 3);
}
// Driver Code
int main()
{
// Given Number N
int N = 19;
if (diffCube(N)) {
cout << "Yes\n";
print(N);
}
else {
cout << "No\n";
}
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to print the two consecutive
// numbers whose difference is N
static void print(int N)
{
// Iterate in the range [0, 10^5]
for (int i = 0; i < 100000; i++)
{
if (Math.pow(i + 1, 3) - Math.pow(i, 3) == N)
{
int j = i + 1;
System.out.println(i + " " + j);
return;
}
}
}
// Function to check if N is a
// perfect cube
static boolean isPerfectSquare(double x)
{
// Find floating point value of
// square root of x.
double sr = Math.sqrt(x);
// If square root is an integer
return ((sr - Math.floor(sr)) == 0);
}
// Function to check whether a number
// can be represented as difference
// of two consecutive cubes
static boolean diffCube(int N)
{
// Check if 12 * N - 3 is a
// perfect square or not
return isPerfectSquare(12 * N - 3);
}
// Driver Code
public static void main(String[] args)
{
// Given Number N
int N = 19;
if (diffCube(N))
{
System.out.println("Yes");
print(N);
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by rock_cool
Python3
# Python3 program for the above approach
import math
# Function to print the two consecutive
# numbers whose difference is N
def printt(N):
# Iterate in the range [0, 10^5]
for i in range(100000):
if (pow(i + 1, 3) - pow(i, 3) == N):
print(i, '', i + 1)
return
# Function to check if N is a
# perfect cube
def isPerfectSquare(x):
# Find floating povalue of
# square root of x.
sr = math.sqrt(x)
# If square root is an integer
return ((sr - math.floor(sr)) == 0)
# Function to check whether a number
# can be represented as difference
# of two consecutive cubes
def diffCube(N):
# Check if 12 * N - 3 is a
# perfect square or not
return isPerfectSquare(12 * N - 3)
# Driver Code
# Given number N
N = 19
if (diffCube(N)):
print("Yes")
printt(N)
else:
print("No")
# This code is contributed by sanjoy_62
C#
// C# program for the above approach
using System;
class GFG{
// Function to print the two consecutive
// numbers whose difference is N
static void print(int N)
{
// Iterate in the range [0, 10^5]
for (int i = 0; i < 100000; i++)
{
if (Math.Pow(i + 1, 3) - Math.Pow(i, 3) == N)
{
int j = i + 1;
Console.WriteLine(i + " " + j);
return;
}
}
}
// Function to check if N is a
// perfect cube
static bool isPerfectSquare(double x)
{
// Find floating point value of
// square root of x.
double sr = Math.Sqrt(x);
// If square root is an integer
return ((sr - Math.Floor(sr)) == 0);
}
// Function to check whether a number
// can be represented as difference
// of two consecutive cubes
static bool diffCube(int N)
{
// Check if 12 * N - 3 is a
// perfect square or not
return isPerfectSquare(12 * N - 3);
}
// Driver Code
public static void Main(String[] args)
{
// Given Number N
int N = 19;
if (diffCube(N))
{
Console.WriteLine("Yes");
print(N);
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript program for the above approach
// Function to print the two consecutive
// numbers whose difference is N
function print(N)
{
// Iterate in the range [0, 10^5]
for(let i = 0; i < 100000; i++)
{
if (parseInt(Math.pow(i + 1, 3), 10) -
parseInt(Math.pow(i, 3), 10) == N)
{
document.write(i + " " + (i + 1));
return;
}
}
}
// Function to check if N is a
// perfect cube
function isPerfectSquare(x)
{
// Find floating point value of
// square root of x.
let sr = Math.sqrt(x);
// If square root is an integer
return ((sr - Math.floor(sr)) == 0);
}
// Function to check whether a number
// can be represented as difference
// of two consecutive cubes
function diffCube(N)
{
// Check if 12 * N - 3 is a
// perfect square or not
return isPerfectSquare(12 * N - 3);
}
// Driver code
// Given Number N
let N = 19;
if (diffCube(N) != 0)
{
document.write("Yes" + "</br>");
print(N);
}
else
{
document.write("No");
}
// This code is contributed by divyeshrabadiya07
</script>
Producción:
Yes 2 3
Complejidad de tiempo: O(N), donde N está en el rango 10 5
Espacio auxiliar: O(1)