Dado un entero K y un rango de números consecutivos [L, R] . La tarea es contar los números del rango dado que tienen raíz digital como K (1 ≤ K ≤ 9). La raíz digital es la suma de los dígitos de un número hasta que se convierte en un número de un solo dígito. Por ejemplo, 256 -> 2 + 5 + 6 = 13 -> 1 + 3 = 4.
Ejemplos:
Entrada: L = 10, R = 22, K = 3
Salida: 2
12 y 21 son los únicos números del rango cuya suma de dígitos es 3.Entrada: L = 100, R = 200, K = 5
Salida: 11
Acercarse:
- Lo primero es tener en cuenta que para cualquier número, la suma de los dígitos es igual al número % 9. Si el resto es 0, entonces la suma de los dígitos es 9.
- Entonces, si K = 9, entonces reemplace K con 0.
- Tarea, ahora es encontrar el conteo de números en el rango L a R con módulo 9 igual a K.
- Divida todo el rango en los máximos grupos posibles de 9 comenzando por L (TotalRange / 9), ya que en cada rango habrá exactamente un número con módulo 9 igual a K.
- Realice un bucle sobre el número restante de elementos de R a R: recuento de elementos restantes y verifique si algún número satisface la condición.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
// Function to return the count
// of required numbers
int countNumbers(int L, int R, int K)
{
if (K == 9)
K = 0;
// Count of numbers present
// in given range
int totalnumbers = R - L + 1;
// Number of groups of 9 elements
// starting from L
int factor9 = totalnumbers / 9;
// Left over elements not covered
// in factor 9
int rem = totalnumbers % 9;
// One Number in each group of 9
int ans = factor9;
// To check if any number in rem
// satisfy the property
for (int i = R; i > R - rem; i--) {
int rem1 = i % 9;
if (rem1 == K)
ans++;
}
return ans;
}
// Driver code
int main()
{
int L = 10;
int R = 22;
int K = 3;
cout << countNumbers(L, R, K);
return 0;
}
Java
// Java implementation of the approach
class GFG {
// Function to return the count
// of required numbers
static int countNumbers(int L, int R, int K) {
if (K == 9) {
K = 0;
}
// Count of numbers present
// in given range
int totalnumbers = R - L + 1;
// Number of groups of 9 elements
// starting from L
int factor9 = totalnumbers / 9;
// Left over elements not covered
// in factor 9
int rem = totalnumbers % 9;
// One Number in each group of 9
int ans = factor9;
// To check if any number in rem
// satisfy the property
for (int i = R; i > R - rem; i--) {
int rem1 = i % 9;
if (rem1 == K) {
ans++;
}
}
return ans;
}
// Driver code
public static void main(String[] args) {
int L = 10;
int R = 22;
int K = 3;
System.out.println(countNumbers(L, R, K));
}
}
/* This code contributed by PrinciRaj1992 */
Python3
# Python3 implementation of the approach # Function to return the count # of required numbers def countNumbers(L, R, K): if (K == 9): K = 0 # Count of numbers present # in given range totalnumbers = R - L + 1 # Number of groups of 9 elements # starting from L factor9 = totalnumbers // 9 # Left over elements not covered # in factor 9 rem = totalnumbers % 9 # One Number in each group of 9 ans = factor9 # To check if any number in rem # satisfy the property for i in range(R, R - rem, -1): rem1 = i % 9 if (rem1 == K): ans += 1 return ans # Driver code L = 10 R = 22 K = 3 print(countNumbers(L, R, K)) # This code is contributed # by mohit kumar
C#
// C# implementation of the approach
using System ;
class GFG
{
// Function to return the count
// of required numbers
static int countNumbers(int L, int R, int K)
{
if (K == 9)
{
K = 0;
}
// Count of numbers present
// in given range
int totalnumbers = R - L + 1;
// Number of groups of 9 elements
// starting from L
int factor9 = totalnumbers / 9;
// Left over elements not covered
// in factor 9
int rem = totalnumbers % 9;
// One Number in each group of 9
int ans = factor9;
// To check if any number in rem
// satisfy the property
for (int i = R; i > R - rem; i--)
{
int rem1 = i % 9;
if (rem1 == K)
{
ans++;
}
}
return ans;
}
// Driver code
public static void Main()
{
int L = 10;
int R = 22;
int K = 3;
Console.WriteLine(countNumbers(L, R, K));
}
}
/* This code is contributed by Ryuga */
PHP
<?php
// PHP implementation of the approach
// Function to return the count
// of required numbers
function countNumbers($L, $R, $K)
{
if ($K == 9)
$K = 0;
// Count of numbers present
// in given range
$totalnumbers = $R - $L + 1;
// Number of groups of 9 elements
// starting from L
$factor9 = intval($totalnumbers / 9);
// Left over elements not covered
// in factor 9
$rem = $totalnumbers % 9;
// One Number in each group of 9
$ans = $factor9;
// To check if any number in rem
// satisfy the property
for ($i = $R; $i > $R - $rem; $i--)
{
$rem1 = $i % 9;
if ($rem1 == $K)
$ans++;
}
return $ans;
}
// Driver code
$L = 10;
$R = 22;
$K = 3;
echo countNumbers($L, $R, $K);
// This code is contributed by Ita_c
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to return the count
// of required numbers
function countNumbers(L, R, K)
{
if (K == 9)
{
K = 0;
}
// Count of numbers present
// in given range
var totalnumbers = R - L + 1;
// Number of groups of 9 elements
// starting from L
var factor9 = totalnumbers / 9;
// Left over elements not covered
// in factor 9
var rem = totalnumbers % 9;
// One Number in each group of 9
var ans = factor9;
// To check if any number in rem
// satisfy the property
for(var i = R; i > R - rem; i--)
{
var rem1 = i % 9;
if (rem1 == K)
{
ans++;
}
}
return ans;
}
// Driver Code
var L = 10;
var R = 22;
var K = 3;
document.write(Math.round(countNumbers(L, R, K)));
// This code is contributed by Ankita saini
</script>
Producción:
2
Complejidad temporal: O(R)
Espacio auxiliar: O(1)