Dado un número entero N mayor que 2, la tarea es imprimir cualquier combinación de a, b y c tal que:
- un + segundo + c = norte
- a, b y c no son divisibles por 3.
Ejemplos:
Input: N = 233 Output: 77 77 79 Input: N = 3 Output: 1 1 1
Un enfoque ingenuo es usar tres bucles y verificar la condición dada.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to print a, b and c
// such that a+b+c=N
#include <bits/stdc++.h>
using namespace std;
// Function to print a, b and c
void printCombination(int n)
{
// first loop
for (int i = 1; i < n; i++) {
// check for 1st number
if (i % 3 != 0) {
// second loop
for (int j = 1; j < n; j++) {
// check for 2nd number
if (j % 3 != 0) {
// third loop
for (int k = 1; k < n; k++) {
// Check for 3rd number
if (k % 3 != 0 && (i + j + k) == n) {
cout << i << " " << j << " " << k;
return;
}
}
}
}
}
}
}
// Driver Code
int main()
{
int n = 233;
printCombination(n);
return 0;
}
Java
// Java program to print a,
// b and c such that a+b+c=N
import java.io.*;
class GFG
{
// Function to print a, b and c
static void printCombination(int n)
{
// first loop
for (int i = 1; i < n; i++)
{
// check for 1st number
if (i % 3 != 0)
{
// second loop
for (int j = 1; j < n; j++)
{
// check for 2nd number
if (j % 3 != 0)
{
// third loop
for (int k = 1; k < n; k++)
{
// Check for 3rd number
if (k % 3 != 0 && (i + j + k) == n)
{
System.out.println( i + " " +
j + " " + k);
return;
}
}
}
}
}
}
}
// Driver Code
public static void main (String[] args)
{
int n = 233;
printCombination(n);
}
}
// This code is contributed
// by anuj_67.
Python3
# Python3 program to print a, b # and c such that a+b+c=N # Function to print a, b and c def printCombination(n): # first loop for i in range(1, n): # check for 1st number if (i % 3 != 0): # second loop for j in range(1, n): # check for 2nd number if (j % 3 != 0): # third loop for k in range(1, n): # Check for 3rd number if (k % 3 != 0 and (i + j + k) == n): print(i, j, k); return; # Driver Code n = 233; printCombination(n); # This code is contributed # by mits
C#
// C# program to print a,
// b and c such that a+b+c=N
class GFG
{
// Function to print a, b and c
static void printCombination(int n)
{
// first loop
for (int i = 1; i < n; i++)
{
// check for 1st number
if (i % 3 != 0)
{
// second loop
for (int j = 1; j < n; j++)
{
// check for 2nd number
if (j % 3 != 0)
{
// third loop
for (int k = 1; k < n; k++)
{
// Check for 3rd number
if (k % 3 != 0 && (i + j + k) == n)
{
System.Console.WriteLine(i + " " +
j + " " + k);
return;
}
}
}
}
}
}
}
// Driver Code
static void Main ()
{
int n = 233;
printCombination(n);
}
}
// This code is contributed
// by mits
PHP
<?php
// PHP program to print a, b and c
// such that a+b+c=N
// Function to print a, b and c
function printCombination($n)
{
// first loop
for ($i = 1; $i < $n; $i++)
{
// check for 1st number
if ($i % 3 != 0)
{
// second loop
for ($j = 1; $j < $n; $j++)
{
// check for 2nd number
if ($j % 3 != 0)
{
// third loop
for ($k = 1; $k < $n; $k++)
{
// Check for 3rd number
if ($k % 3 != 0 &&
($i + $j + $k) == $n)
{
echo $i , " " , $j , " " , $k;
return;
}
}
}
}
}
}
}
// Driver Code
$n = 233;
printCombination($n);
// This code is contributed
// inder_verma
?>
Javascript
<script>
// Javascript program to print a, b and c
// such that a+b+c=N
// Function to print a, b and c
function printCombination(n)
{
// first loop
for (let i = 1; i < n; i++) {
// check for 1st number
if (i % 3 != 0) {
// second loop
for (let j = 1; j < n; j++) {
// check for 2nd number
if (j % 3 != 0) {
// third loop
for (let k = 1; k < n; k++) {
// Check for 3rd number
if (k % 3 != 0 && (i + j + k) == n) {
document.write(i + " " + j + " " + k);
return;
}
}
}
}
}
}
}
// Driver Code
let n = 233;
printCombination(n);
// This code is contributed by rishavmahato348.
</script>
Producción:
1 2 230
Complejidad de tiempo : O(N 3 ), ya que estamos usando un bucle para atravesar N 3 veces.
Espacio auxiliar : O(1), ya que no estamos utilizando ningún espacio adicional.
Enfoque eficiente:
- Considere el 1 como uno de los tres números.
- Los otros dos números serán:
- (2, n-3) si n-2 es divisible por 3.
- De lo contrario (1, n-2) si n-2 no es divisible por 3.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to print a, b and c
// such that a+b+c=N
#include <bits/stdc++.h>
using namespace std;
// Function to print a, b and c
void printCombination(int n)
{
cout << 1 << " ";
// check if n-2 is divisible
// by 3 or not
if ((n - 2) % 3 == 0)
cout << 2 << " " << n - 3;
else
cout << 1 << " " << n - 2;
}
// Driver Code
int main()
{
int n = 233;
printCombination(n);
return 0;
}
Java
// Java program to print a, b
// and c such that a+b+c=N
class GFG
{
// Function to print a, b and c
static void printCombination(int n)
{
System.out.print(1 + " ");
// check if n-2 is divisible
// by 3 or not
if ((n - 2) % 3 == 0)
System.out.print(2 + " " + (n - 3));
else
System.out.print(1 + " " + (n - 2));
}
// Driver Code
public static void main(String[] args)
{
int n = 233;
printCombination(n);
}
}
// This code is contributed by mits
Python3
# Python3 program to print a, b and c
# such that a+b+c=N
# Function to print a, b and c
def printCombination(n):
print("1 ",end="");
# check if n-2 is divisible
# by 3 or not
if ((n - 2) % 3 == 0):
print("2",n - 3,end="");
else:
print("1",(n - 2),end="");
# Driver code
if __name__=='__main__':
n = 233;
printCombination(n);
# This code is contributed by mits
C#
// C# program to print a, b
// and c such that a+b+c=N
class GFG
{
// Function to print a, b and c
static void printCombination(int n)
{
System.Console.Write(1 + " ");
// check if n-2 is divisible
// by 3 or not
if ((n - 2) % 3 == 0)
System.Console.Write(2 + " " + (n - 3));
else
System.Console.Write(1 + " " + (n - 2));
}
// Driver Code
static void Main()
{
int n = 233;
printCombination(n);
}
}
// This code is contributed by mits
PHP
<?php
// PHP program to print a, b and c
// such that a+b+c=N
// Function to print a, b and c
function printCombination($n)
{
echo "1 ";
// check if n-2 is divisible
// by 3 or not
if (($n - 2) % 3 == 0)
echo "2 " . ($n - 3);
else
echo "1 " . ($n - 2);
}
// Driver code
$n = 233;
printCombination($n);
// This code is contributed by mits
?>
Javascript
<script>
// Javascript program to print a, b and c
// such that a+b+c=N
// Function to print a, b and c
function printCombination(n)
{
document.write(1 + " ");
// Check if n-2 is divisible
// by 3 or not
if ((n - 2) % 3 == 0)
document.write(2 + " " + (n - 3));
else
document.write(1 + " " + (n - 2));
}
// Driver Code
let n = 233;
printCombination(n);
// This code is contributed by subhammahato348
</script>
Producción:
1 2 230
Complejidad de tiempo : O (1), ya que no estamos usando ningún bucle o recursión para atravesar.
Espacio auxiliar : O(1), ya que no estamos utilizando ningún espacio adicional.