Dados n dígitos distintos (de 0 a 9), encuentre la suma de todos los números de n dígitos que se pueden formar usando estos dígitos. Se supone que se permiten los números formados con 0 a la izquierda.
Ejemplo:
Input: 1 2 3 Output: 1332 Explanation Numbers Formed: 123 , 132 , 312 , 213, 231 , 321 123 + 132 + 312 + 213 + 231 + 321 = 1332
Los números totales que se pueden formar usando n dígitos es el número total de permutaciones de n dígitos, es decir, factorial(n). Ahora, dado que el número formado es un número de n dígitos, cada dígito aparecerá factorial (n)/n veces en cada posición desde el dígito menos significativo hasta el dígito más significativo. Por lo tanto, suma de dígitos en una posición = (suma de todos los dígitos) * (factorial(n)/n).
Considering the example digits as 1 2 3 factorial(3)/3 = 2 Sum of digits at least significant digit = (1 + 2 + 3) * 2 = 12 Similarly sum of digits at tens, hundreds place is 12. (This sum will contribute as 12 * 100) Similarly sum of digits at tens, thousands place is 12. (This sum will contribute as 12 * 1000) Required sum of all numbers = 12 + (10 * 12) + (100 * 12) = 1332
Implementación:
C++
// C++ program to find sun of numbers formed
// by all permutations of given set of digits
#include<stdio.h>
// function to calculate factorial of a number
int factorial(int n)
{
int f = 1;
if (n==0||n==1)
return 1;
for (int i=2; i<=n; i++)
f = f*i;
return f;
}
// Function to calculate sum of all numbers
int getSum(int arr[],int n)
{
// calculate factorial
int fact = factorial(n);
// sum of all the given digits at different
// positions is same and is going to be stored
// in digitsum.
int digitsum = 0;
for (int i=0; i<n; i++)
digitsum += arr[i];
digitsum *= (fact/n);
// Compute result (sum of all the numbers)
int res = 0;
for (int i=1, k=1; i<=n; i++)
{
res += (k*digitsum);
k = k*10;
}
return res;
}
// Driver program to test above function
int main()
{
// n distinct digits
int arr[] = {1, 2, 3};
int n = sizeof(arr)/sizeof(arr[0]);
// Print sum of all the numbers formed
printf("%d", getSum(arr, n));
return 0;
}
Java
// Java program to find sum
// of numbers formed by all
// permutations of given set
// of digits
import java.io.*;
class GFG
{
// function to calculate
// factorial of a number
static int factorial(int n)
{
int f = 1;
if (n == 0|| n == 1)
return 1;
for (int i = 2; i <= n; i++)
f = f * i;
return f;
}
// Function to calculate
// sum of all numbers
static int getSum(int arr[], int n)
{
// calculate factorial
int fact = factorial(n);
// sum of all the given
// digits at different
// positions is same and
// is going to be stored
// in digitsum.
int digitsum = 0;
for (int i = 0; i < n; i++)
digitsum += arr[i];
digitsum *= (fact / n);
// Compute result (sum
// of all the numbers)
int res = 0;
for (int i = 1, k = 1;
i <= n; i++)
{
res += (k * digitsum);
k = k * 10;
}
return res;
}
// Driver Code
public static void main (String[] args)
{
// n distinct digits
int arr[] = {1, 2, 3};
int n = arr.length;
// Print sum of all
// the numbers formed
System.out.println(getSum(arr, n));
}
}
// This code is contributed
// by ajit
Python3
# Python3 program to find sun of # numbers formed by all permutations # of given set of digits # function to calculate factorial # of a number def factorial(n): f = 1 if (n == 0 or n == 1): return 1 for i in range(2, n + 1): f = f * i return f # Function to calculate sum # of all numbers def getSum(arr, n): # calculate factorial fact = factorial(n) # sum of all the given digits at # different positions is same and # is going to be stored in digitsum. digitsum = 0 for i in range(n): digitsum += arr[i] digitsum *= (fact // n) # Compute result (sum of # all the numbers) res = 0 i = 1 k = 1 while i <= n : res += (k * digitsum) k = k * 10 i += 1 return res # Driver Code if __name__ == "__main__": # n distinct digits arr = [1, 2, 3] n = len(arr) # Print sum of all the numbers formed print(getSum(arr, n)) # This code is contributed by ita_c
C#
// C# program to find sum
// of numbers formed by all
// permutations of given set
// of digits
using System;
class GFG
{
// function to calculate
// factorial of a number
static int factorial(int n)
{
int f = 1;
if (n == 0|| n == 1)
return 1;
for (int i = 2; i <= n; i++)
f = f * i;
return f;
}
// Function to calculate
// sum of all numbers
static int getSum(int []arr,
int n)
{
// calculate factorial
int fact = factorial(n);
// sum of all the given
// digits at different
// positions is same and
// is going to be stored
// in digitsum.
int digitsum = 0;
for (int i = 0; i < n; i++)
digitsum += arr[i];
digitsum *= (fact / n);
// Compute result (sum
// of all the numbers)
int res = 0;
for (int i = 1, k = 1;
i <= n; i++)
{
res += (k * digitsum);
k = k * 10;
}
return res;
}
// Driver Code
static public void Main ()
{
// n distinct digits
int []arr = {1, 2, 3};
int n = arr.Length;
// Print sum of all
// the numbers formed
Console.WriteLine(getSum(arr, n));
}
}
// This code is contributed
// by akt_mit
PHP
<?php
// PHP program to find sum
// of numbers formed by all
// permutations of given set
// of digits function to
// calculate factorial of a number
function factorial($n)
{
$f = 1;
if ($n == 0||$n == 1)
return 1;
for ($i = 2; $i <= $n; $i++)
$f = $f * $i;
return $f;
}
// Function to calculate
// sum of all numbers
function getSum($arr,$n)
{
// calculate factorial
$fact = factorial($n);
// sum of all the given
// digits at different
// positions is same and
// is going to be stored
// in digitsum.
$digitsum = 0;
for ($i = 0; $i < $n; $i++)
$digitsum += $arr[$i];
$digitsum *= ($fact / $n);
// Compute result (sum
// of all the numbers)
$res = 0;
for ($i = 1, $k = 1; $i <= $n; $i++)
{
$res += ($k * $digitsum);
$k = $k * 10;
}
return $res;
}
// Driver Code
// n distinct digits
$arr = array(1, 2, 3);
$n = sizeof($arr);
// Print sum of all
// the numbers formed
echo getSum($arr, $n);
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript program to find sum of
// numbers formed by all permutations
// of given set of digits
// Function to calculate
// factorial of a number
function factorial(n)
{
let f = 1;
if (n == 0 || n == 1)
return 1;
for (let i = 2; i <= n; i++)
f = f * i;
return f;
}
// Function to calculate
// sum of all numbers
function getSum(arr, n)
{
// Calculate factorial
let fact = factorial(n);
// Sum of all the given
// digits at different
// positions is same and
// is going to be stored
// in digitsum.
let digitsum = 0;
for (let i = 0; i < n; i++)
digitsum += arr[i];
digitsum *= (fact / n);
// Compute result (sum
// of all the numbers)
let res = 0;
for(let i = 1, k = 1;
i <= n; i++)
{
res += (k * digitsum);
k = k * 10;
}
return res;
}
// Driver code
// n distinct digits
let arr = [1, 2, 3];
let n = arr.length;
// Print sum of all
// the numbers formed
document.write(getSum(arr, n));
// This code is contributed by susmitakundugoaldanga
</script>
Producción
1332
Complejidad temporal: O(n)
Espacio auxiliar: O(1)
Este artículo es una contribución de Harsh Agarwal . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA