Dado un entero positivo N y un dígito D , la tarea es comprobar si N puede representarse como una suma de enteros positivos que contengan el dígito D al menos una vez. Si es posible representar N en dicho formato, imprima «Sí» . De lo contrario, escriba “No” .
Ejemplos:
Entrada: N = 24, D = 7
Salida: Sí
Explicación: El valor 24 se puede representar como 17 + 7, ambos con el dígito 7.Entrada: N = 27 D = 2
Salida: Sí
Enfoque: Siga los pasos para resolver el problema:
- Compruebe si la N dada contiene el dígito D o no. Si se encuentra que es cierto , escriba «Sí» .
- De lo contrario, itere hasta que N sea mayor que 0 y realice los siguientes pasos:
- Restando el valor D del valor de N .
- Compruebe si el valor actualizado de N contiene el dígito D o no. Si se determina que es cierto , escriba «Sí» y salga del bucle .
- Después de completar los pasos anteriores, si ninguna de las condiciones anteriores cumple, imprima «No» .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <iostream>
using namespace std;
// Function to check if N contains
// digit D in it
bool findDigit(int N, int D)
{
// Iterate until N is positive
while (N > 0) {
// Find the last digit
int a = N % 10;
// If the last digit is the
// same as digit D
if (a == D) {
return true;
}
N /= 10;
}
// Return false
return false;
}
// Function to check if the value of
// N can be represented as sum of
// integers having digit d in it
bool check(int N, int D)
{
// Iterate until N is positive
while (N > 0) {
// Check if N contains digit
// D or not
if (findDigit(N, D) == true) {
return true;
}
// Subtracting D from N
N -= D;
}
// Return false
return false;
}
// Driver Code
int main()
{
int N = 24;
int D = 7;
if (check(N, D)) {
cout << "Yes";
}
else {
cout << "No";
}
return 0;
}
Java
// Java approach for the above approach
import java.util.*;
class GFG{
// Function to check if N contains
// digit D in it
static boolean findDigit(int N, int D)
{
// Iterate until N is positive
while (N > 0)
{
// Find the last digit
int a = N % 10;
// If the last digit is the
// same as digit D
if (a == D)
{
return true;
}
N /= 10;
}
// Return false
return false;
}
// Function to check if the value of
// N can be represented as sum of
// integers having digit d in it
static boolean check(int N, int D)
{
// Iterate until N is positive
while (N > 0)
{
// Check if N contains digit
// D or not
if (findDigit(N, D) == true)
{
return true;
}
// Subtracting D from N
N -= D;
}
// Return false
return false;
}
// Driver Code
public static void main(String[] args)
{
int N = 24;
int D = 7;
if (check(N, D))
{
System.out.print("Yes");
}
else
{
System.out.print("No");
}
}
}
// This code is contributed by sanjoy_62
Python3
# Python3 program for the above approach
# Function to check if N contains
# digit D in it
def findDigit(N, D):
# Iterate until N is positive
while (N > 0):
# Find the last digit
a = N % 10
# If the last digit is the
# same as digit D
if (a == D):
return True
N /= 10
# Return false
return False
# Function to check if the value of
# N can be represented as sum of
# integers having digit d in it
def check(N, D):
# Iterate until N is positive
while (N > 0):
# Check if N contains digit
# D or not
if (findDigit(N, D) == True):
return True
# Subtracting D from N
N -= D
# Return false
return False
# Driver Code
if __name__ == '__main__':
N = 24
D = 7
if (check(N, D)):
print("Yes")
else:
print("No")
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG{
// Function to check if N contains
// digit D in it
static bool findDigit(int N, int D)
{
// Iterate until N is positive
while (N > 0)
{
// Find the last digit
int a = N % 10;
// If the last digit is the
// same as digit D
if (a == D)
{
return true;
}
N /= 10;
}
// Return false
return false;
}
// Function to check if the value of
// N can be represented as sum of
// integers having digit d in it
static bool check(int N, int D)
{
// Iterate until N is positive
while (N > 0)
{
// Check if N contains digit
// D or not
if (findDigit(N, D) == true)
{
return true;
}
// Subtracting D from N
N -= D;
}
// Return false
return false;
}
// Driver Code
public static void Main()
{
int N = 24;
int D = 7;
if (check(N, D))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by code_hunt.
Javascript
<script>
// javascript program for the above approach
// Function to check if N contains
// digit D in it
function findDigit(N, D)
{
// Iterate until N is positive
while (N > 0)
{
// Find the last digit
let a = N % 10;
// If the last digit is the
// same as digit D
if (a == D)
{
return true;
}
N = Math.floor(N / 10);
}
// Return false
return false;
}
// Function to check if the value of
// N can be represented as sum of
// integers having digit d in it
function check(N, D)
{
// Iterate until N is positive
while (N > 0)
{
// Check if N contains digit
// D or not
if (findDigit(N, D) == true)
{
return true;
}
// Subtracting D from N
N -= D;
}
// Return false
return false;
}
// Driver Code
let N = 24;
let D = 7;
if (check(N, D))
{
document.write("Yes");
}
else
{
document.write("No");
}
</script>
Producción:
Yes
Complejidad de tiempo: O(N)
Complejidad de espacio: O(1)
Publicación traducida automáticamente
Artículo escrito por RohitOberoi y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA