Dados tres números enteros N , K y L . La tarea es encontrar el promedio de los primeros K dígitos y los últimos L dígitos del número N dado sin que ningún dígito se superponga.
Ejemplos:
Entrada: N = 123456, K = 2, L = 3
Salida: 3.0 La
suma de los primeros K dígitos será 1 + 2 = 3 La
suma de los últimos L dígitos será 4 + 5 + 6 = 15
Promedio = (3 + 15) / (2 + 3) = 18 / 5 = 3
Entrada: N = 456966, K = 1, L = 1
Salida: 5,0
Enfoque: si el recuento de dígitos en n es menor que (K + L) , entonces no es posible encontrar el promedio sin que los dígitos se superpongan e imprimir -1 en ese caso. Si ese no es el caso, encuentre la suma de los últimos L dígitos de N y guárdela en una variable, digamos sum1 , luego encuentre la suma de los primeros K dígitos de N y guárdela en sum2 . Ahora, imprima el promedio como (sum1 + sum2) / (K + L) .
A continuación se muestra la implementación del enfoque anterior:
C++
// implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count
// of digits in num
int countDigits(int num)
{
int cnt = 0;
while (num > 0)
{
cnt++;
num /= 10;
}
return cnt;
}
// Function to return the sum
// of first n digits of num
int sumFromStart(int num, int n, int rem)
{
// Remove the unnecessary digits
num /= ((int)pow(10, rem));
int sum = 0;
while (num > 0)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to return the sum
// of the last n digits of num
int sumFromEnd(int num, int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
float getAverage(int n, int k, int l)
{
// If the average can't be calculated without
// using the same digit more than once
int totalDigits = countDigits(n);
if (totalDigits < (k + l))
return -1;
// Sum of the last l digits of n
int sum1 = sumFromEnd(n, l);
// Sum of the first k digits of n
// (totalDigits - k) must be removed from the
// end of the number to get the remaining
// k digits from the beginning
int sum2 = sumFromStart(n, k, totalDigits - k);
// Return the average
return ((float)(sum1 + sum2) /
(float)(k + l));
}
// Driver code
int main()
{
int n = 123456, k = 2, l = 3;
cout << getAverage(n, k, l);
return 0;
}
// This code is contributed by PrinciRaj1992
Java
// Java implementation of the approach
class GFG {
// Function to return the count
// of digits in num
public static int countDigits(int num)
{
int cnt = 0;
while (num > 0) {
cnt++;
num /= 10;
}
return cnt;
}
// Function to return the sum
// of first n digits of num
public static int sumFromStart(int num, int n, int rem)
{
// Remove the unnecessary digits
num /= ((int)Math.pow(10, rem));
int sum = 0;
while (num > 0) {
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to return the sum
// of the last n digits of num
public static int sumFromEnd(int num, int n)
{
int sum = 0;
for (int i = 0; i < n; i++) {
sum += (num % 10);
num /= 10;
}
return sum;
}
public static float getAverage(int n, int k, int l)
{
// If the average can't be calculated without
// using the same digit more than once
int totalDigits = countDigits(n);
if (totalDigits < (k + l))
return -1;
// Sum of the last l digits of n
int sum1 = sumFromEnd(n, l);
// Sum of the first k digits of n
// (totalDigits - k) must be removed from the
// end of the number to get the remaining
// k digits from the beginning
int sum2 = sumFromStart(n, k, totalDigits - k);
// Return the average
return ((float)(sum1 + sum2) / (float)(k + l));
}
// Driver code
public static void main(String args[])
{
int n = 123456, k = 2, l = 3;
System.out.print(getAverage(n, k, l));
}
}
Python3
# implementation of the approach from math import pow # Function to return the count # of digits in num def countDigits(num): cnt = 0 while (num > 0): cnt += 1 num //= 10 return cnt # Function to return the sum # of first n digits of num def sumFromStart(num, n, rem): # Remove the unnecessary digits num //= pow(10, rem) sum = 0 while (num > 0): sum += (num % 10) num //= 10 return sum # Function to return the sum # of the last n digits of num def sumFromEnd(num, n): sum = 0 for i in range(n): sum += (num % 10) num //= 10 return sum def getAverage(n, k, l): # If the average can't be calculated without # using the same digit more than once totalDigits = countDigits(n) if (totalDigits < (k + l)): return -1 # Sum of the last l digits of n sum1 = sumFromEnd(n, l) # Sum of the first k digits of n # (totalDigits - k) must be removed from the # end of the number to get the remaining # k digits from the beginning sum2 = sumFromStart(n, k, totalDigits - k) # Return the average return (sum1 + sum2) / (k + l) # Driver code if __name__ == '__main__': n = 123456 k = 2 l = 3 print(getAverage(n, k, l)) # This code is contributed by # Surendra_Gangwar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count
// of digits in num
public static int countDigits(int num)
{
int cnt = 0;
while (num > 0)
{
cnt++;
num /= 10;
}
return cnt;
}
// Function to return the sum
// of first n digits of num
public static int sumFromStart(int num,
int n, int rem)
{
// Remove the unnecessary digits
num /= ((int)Math.Pow(10, rem));
int sum = 0;
while (num > 0)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to return the sum
// of the last n digits of num
public static int sumFromEnd(int num, int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
public static float getAverage(int n, int k, int l)
{
// If the average can't be calculated without
// using the same digit more than once
int totalDigits = countDigits(n);
if (totalDigits < (k + l))
return -1;
// Sum of the last l digits of n
int sum1 = sumFromEnd(n, l);
// Sum of the first k digits of n
// (totalDigits - k) must be removed from the
// end of the number to get the remaining
// k digits from the beginning
int sum2 = sumFromStart(n, k, totalDigits - k);
// Return the average
return ((float)(sum1 + sum2) /
(float)(k + l));
}
// Driver code
public static void Main(String []args)
{
int n = 123456, k = 2, l = 3;
Console.WriteLine(getAverage(n, k, l));
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// javascript implementation of the approach
// Function to return the count
// of digits in num
function countDigits(num)
{
var cnt = 0;
while (num > 0)
{
cnt++;
num = parseInt(num/10);
}
return cnt;
}
// Function to return the sum
// of first n digits of num
function sumFromStart(num, n, rem)
{
// Remove the unnecessary digits
num = (parseInt( num/Math.pow(10, rem)));
var sum = 0;
while (num > 0)
{
sum += (num % 10);
num = parseInt(num/10);
}
return sum;
}
// Function to return the sum
// of the last n digits of num
function sumFromEnd(num , n)
{
var sum = 0;
for (i = 0; i < n; i++)
{
sum += (num % 10);
num = parseInt(num/10);
}
return sum;
}
function getAverage(n , k , l) {
// If the average can't be calculated without
// using the same digit more than once
var totalDigits = countDigits(n);
if (totalDigits < (k + l))
return -1;
// Sum of the last l digits of n
var sum1 = sumFromEnd(n, l);
// Sum of the first k digits of n
// (totalDigits - k) must be removed from the
// end of the number to get the remaining
// k digits from the beginning
var sum2 = sumFromStart(n, k, totalDigits - k);
// Return the average
return ( (sum1 + sum2) / (k + l));
}
// Driver code
var n = 123456, k = 2, l = 3;
document.write(getAverage(n, k, l));
// This code is contributed by Rajput-Ji
</script>
Producción:
3.6
Publicación traducida automáticamente
Artículo escrito por mayanksrivastavaramji y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA