Dado N, cuente todos los ‘a’ y ‘b’ que satisfacen la condición a^3 + b^3 = N.
Ejemplos:
Input : N = 9 Output : 2 1^3 + 2^3 = 9 2^3 + 1^3 = 9 Input : N = 28 Output : 2 1^3 + 3^3 = 28 3^3 + 1^3 = 28
Nota: (a, b) y (b, a) deben considerarse como dos pares diferentes.
Preguntado en: Adobe
Implementation:
Traverse numbers from 1 to cube root of N.
a) Subtract cube of current number from
N and check if their difference is a
perfect cube or not.
i) If perfect cube then increment count.
2- Return count.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to count pairs whose sum
// cubes is N
#include<bits/stdc++.h>
using namespace std;
// Function to count the pairs satisfying
// a ^ 3 + b ^ 3 = N
int countPairs(int N)
{
int count = 0;
// Check for each number 1 to cbrt(N)
for (int i = 1; i <= cbrt(N); i++)
{
// Store cube of a number
int cb = i*i*i;
// Subtract the cube from given N
int diff = N - cb;
// Check if the difference is also
// a perfect cube
int cbrtDiff = cbrt(diff);
// If yes, then increment count
if (cbrtDiff*cbrtDiff*cbrtDiff == diff)
count++;
}
// Return count
return count;
}
// Driver program
int main()
{
// Loop to Count no. of pairs satisfying
// a ^ 3 + b ^ 3 = i for N = 1 to 10
for (int i = 1; i<= 10; i++)
cout << "For n = " << i << ", "
<< countPairs(i) <<" pair exists\n";
return 0;
}
Java
// Java program to count pairs whose sum
// cubes is N
class Test
{
// method to count the pairs satisfying
// a ^ 3 + b ^ 3 = N
static int countPairs(int N)
{
int count = 0;
// Check for each number 1 to cbrt(N)
for (int i = 1; i <= Math.cbrt(N); i++)
{
// Store cube of a number
int cb = i*i*i;
// Subtract the cube from given N
int diff = N - cb;
// Check if the difference is also
// a perfect cube
int cbrtDiff = (int) Math.cbrt(diff);
// If yes, then increment count
if (cbrtDiff*cbrtDiff*cbrtDiff == diff)
count++;
}
// Return count
return count;
}
// Driver method
public static void main(String args[])
{
// Loop to Count no. of pairs satisfying
// a ^ 3 + b ^ 3 = i for N = 1 to 10
for (int i = 1; i<= 10; i++)
System.out.println("For n = " + i + ", " +
+ countPairs(i) + " pair exists");
}
}
Python 3
# Python 3 program to count pairs
# whose sum cubes is N
import math
# Function to count the pairs
# satisfying a ^ 3 + b ^ 3 = N
def countPairs(N):
count = 0
# Check for each number 1
# to cbrt(N)
for i in range(1, int(math.pow(N, 1/3) + 1)):
# Store cube of a number
cb = i * i * i
# Subtract the cube from given N
diff = N - cb
# Check if the difference is also
# a perfect cube
cbrtDiff = int(math.pow(diff, 1/3))
# If yes, then increment count
if (cbrtDiff * cbrtDiff * cbrtDiff == diff):
count += 1
# Return count
return count
# Driver program
# Loop to Count no. of pairs satisfying
# a ^ 3 + b ^ 3 = i for N = 1 to 10
for i in range(1, 11):
print('For n = ', i, ', ', countPairs(i),
' pair exists')
# This code is contributed by Smitha.
C#
// C# program to count pairs whose sum
// cubes is N
using System;
class Test
{
// method to count the pairs satisfying
// a ^ 3 + b ^ 3 = N
static int countPairs(int N)
{
int count = 0;
// Check for each number 1 to cbrt(N)
for (int i = 1; i <= Math.Pow(N,(1.0/3.0)); i++)
{
// Store cube of a number
int cb = i*i*i;
// Subtract the cube from given N
int diff = N - cb;
// Check if the difference is also
// a perfect cube
int cbrtDiff = (int) Math.Pow(diff,(1.0/3.0));
// If yes, then increment count
if (cbrtDiff*cbrtDiff*cbrtDiff == diff)
count++;
}
// Return count
return count;
}
// Driver method
public static void Main()
{
// Loop to Count no. of pairs satisfying
// a ^ 3 + b ^ 3 = i for N = 1 to 10
for (int i = 1; i<= 10; i++)
Console.Write("For n = " + i + ", " +
+ countPairs(i) + " pair exists"+"\n");
}
}
PHP
<?php
// PHP program to count pairs
// whose sum cubes is N
// Function to count the pairs
// satisfying a ^ 3 + b ^ 3 = N
function countPairs($N)
{
$count = 0;
// Check for each number
// 1 to cbrt(N)
for ($i = 1;
$i <= (int)pow($N, 1 / 3); $i++)
{
// Store cube of a number
$cb = $i * $i * $i;
// Subtract the cube from
// given N
$diff = ($N - $cb);
// Check if the difference is
// also a perfect cube
$cbrtDiff = (int)pow($diff, 1 / 3);
// If yes, then increment count
if ($cbrtDiff * $cbrtDiff *
$cbrtDiff == $diff)
$count++;
}
// Return count
return $count;
}
// Driver Code
// Loop to Count no. of pairs
// satisfying a ^ 3 + b ^ 3 = i
// for N = 1 to 10
for ($i = 1; $i<= 10; $i++)
echo "For n = " , $i , ", ",
countPairs($i) ," pair exists\n";
// This code is contributed by jit_t
?>
Javascript
<script>
// Javascript program to count pairs whose sum cubes is N
// method to count the pairs satisfying
// a ^ 3 + b ^ 3 = N
function countPairs(N)
{
let count = 0;
// Check for each number 1 to cbrt(N)
for (let i = 1; i <= parseInt(Math.pow(N,(1.0/3.0)), 10); i++)
{
// Store cube of a number
let cb = i*i*i;
// Subtract the cube from given N
let diff = N - cb;
// Check if the difference is also
// a perfect cube
let cbrtDiff = parseInt(Math.pow(diff,(1.0/3.0)), 10);
// If yes, then increment count
if (cbrtDiff*cbrtDiff*cbrtDiff == diff)
count++;
}
// Return count
return count;
}
// Loop to Count no. of pairs satisfying
// a ^ 3 + b ^ 3 = i for N = 1 to 10
for (let i = 1; i<= 10; i++)
document.write("For n = " + i + ", " + countPairs(i) + " pair exists"+"</br>");
</script>
Producción:
For n= 1, 1 pair exists For n= 2, 1 pair exists For n= 3, 0 pair exists For n= 4, 0 pair exists For n= 5, 0 pair exists For n= 6, 0 pair exists For n= 7, 0 pair exists For n= 8, 1 pair exists For n= 9, 2 pair exists For n= 10, 0 pair exists
Referencia: https://www.careercup.com/question?id=5954491572551680
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA