Dados dos números enteros N y X , la tarea es encontrar el recuento mínimo de números enteros con suma N y que tengan el dígito unitario X. Si no existe tal representación, imprima -1 .
Ejemplos:
Entrada: N = 38, X = 9
Salida: 2
Explicación:
Se requiere un mínimo de dos números enteros con dígito de unidad como X para representar como una suma igual a 38.
38 = 19 + 19 o 38 = 29 + 9
Entrada: N = 6, X = 4
Salida: -1
Explicación:
No existe tal representación de
Enfoque:
siga los pasos a continuación para resolver el problema:
- Obtener la cifra unitaria de N y comprobar si se consigue mediante la suma de números cuyas cifras unitarias sean X.
- Si es posible, compruebe si N ? X * (Número mínimo de veces que se debe sumar un número con el dígito unitario X para obtener la suma N ) .
- Si se cumple la condición anterior, imprima la cantidad mínima de veces que se debe agregar un número con el dígito de la unidad X para obtener la suma N. De lo contrario, imprima -1.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
int check(int unit_digit, int X)
{
int times, digit;
// Calculate the number of
// additions required to get unit
// digit of N
for (int times = 1; times <= 10;
times++) {
digit = (X * times) % 10;
if (digit == unit_digit)
return times;
}
// If unit digit of N
// cannot be obtained
return -1;
}
// Function to return the minimum
// number required to represent N
int getNum(int N, int X)
{
int unit_digit;
// Stores unit digit of N
unit_digit = N % 10;
// Stores minimum addition
// of X required to
// obtain unit digit of N
int times = check(unit_digit, X);
// If unit digit of N
// cannot be obtained
if (times == -1)
return times;
// Otherwise
else {
// If N is greater than
// or equal to (X*times)
if (N >= (times * X))
// Minimum count of numbers
// that needed to represent N
return times;
// Representation not
// possible
else
return -1;
}
}
// Driver Code
int main()
{
int N = 58, X = 7;
cout << getNum(N, X) << endl;
return 0;
}
Java
// Java Program to implement
// the above approach
class GFG{
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
static int check(int unit_digit, int X)
{
int times, digit;
// Calculate the number of
// additions required to get unit
// digit of N
for (times = 1; times <= 10;
times++)
{
digit = (X * times) % 10;
if (digit == unit_digit)
return times;
}
// If unit digit of N
// cannot be obtained
return -1;
}
// Function to return the minimum
// number required to represent N
static int getNum(int N, int X)
{
int unit_digit;
// Stores unit digit of N
unit_digit = N % 10;
// Stores minimum addition
// of X required to
// obtain unit digit of N
int times = check(unit_digit, X);
// If unit digit of N
// cannot be obtained
if (times == -1)
return times;
// Otherwise
else
{
// If N is greater than
// or equal to (X*times)
if (N >= (times * X))
// Minimum count of numbers
// that needed to represent N
return times;
// Representation not
// possible
else
return -1;
}
}
// Driver Code
public static void main(String []args)
{
int N = 58, X = 7;
System.out.println( getNum(N, X));
}
}
// This code is contributed by Ritik Bansal
Python3
# Python3 program to implement # the above approach # Function to calculate and return # the minimum number of times a number # with unit digit X needs to be added # to get a sum N def check(unit_digit, X): # Calculate the number of additions # required to get unit digit of N for times in range(1, 11): digit = (X * times) % 10 if (digit == unit_digit): return times # If unit digit of N # cannot be obtained return -1 # Function to return the minimum # number required to represent N def getNum(N, X): # Stores unit digit of N unit_digit = N % 10 # Stores minimum addition # of X required to # obtain unit digit of N times = check(unit_digit, X) # If unit digit of N # cannot be obtained if (times == -1): return times # Otherwise else: # If N is greater than # or equal to (X*times) if (N >= (times * X)): # Minimum count of numbers # that needed to represent N return times # Representation not # possible else: return -1 # Driver Code N = 58 X = 7 print(getNum(N, X)) # This code is contributed by Sanjit_Prasad
C#
// C# Program to implement
// the above approach
using System;
class GFG{
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
static int check(int unit_digit, int X)
{
int times, digit;
// Calculate the number of
// additions required to get unit
// digit of N
for (times = 1; times <= 10;
times++)
{
digit = (X * times) % 10;
if (digit == unit_digit)
return times;
}
// If unit digit of N
// cannot be obtained
return -1;
}
// Function to return the minimum
// number required to represent N
static int getNum(int N, int X)
{
int unit_digit;
// Stores unit digit of N
unit_digit = N % 10;
// Stores minimum addition
// of X required to
// obtain unit digit of N
int times = check(unit_digit, X);
// If unit digit of N
// cannot be obtained
if (times == -1)
return times;
// Otherwise
else
{
// If N is greater than
// or equal to (X*times)
if (N >= (times * X))
// Minimum count of numbers
// that needed to represent N
return times;
// Representation not
// possible
else
return -1;
}
}
// Driver Code
public static void Main()
{
int N = 58, X = 7;
Console.Write(getNum(N, X));
}
}
// This code is contributed by Code_Mech
Javascript
<script>
// Javascript program implementation
// of the approach
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
function check(unit_digit, X)
{
let times, digit;
// Calculate the number of
// additions required to get unit
// digit of N
for (times = 1; times <= 10;
times++)
{
digit = (X * times) % 10;
if (digit == unit_digit)
return times;
}
// If unit digit of N
// cannot be obtained
return -1;
}
// Function to return the minimum
// number required to represent N
function getNum(N, X)
{
let unit_digit;
// Stores unit digit of N
unit_digit = N % 10;
// Stores minimum addition
// of X required to
// obtain unit digit of N
let times = check(unit_digit, X);
// If unit digit of N
// cannot be obtained
if (times == -1)
return times;
// Otherwise
else
{
// If N is greater than
// or equal to (X*times)
if (N >= (times * X))
// Minimum count of numbers
// that needed to represent N
return times;
// Representation not
// possible
else
return -1;
}
}
// Driver Code
let N = 58, X = 7;
document.write( getNum(N, X));
// This code is contributed by splevel62.
</script>
Producción:
4
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)