Dado un número N , imprime todos los Números Fuertes menores o iguales a N .
Número fuerte es un número especial cuya suma del factorial de dígitos es igual al número original.
Por ejemplo: 145 es un número fuerte. Desde, 1! + 4! + 5! = 145.
Ejemplos:
Entrada: N = 100
Salida: 1 2
Explicación:
¡Solo 1 y 2 son los números fuertes del 1 al 100 porque
1! = 1 y
2! = 2Entrada: N = 1000
Salida: 1 2 145
Explicación:
¡Solo 1, 2 y 145 son los números fuertes del 1 al 1000 porque
1! = 1,
2! = 2, y
(1! + 4! + 5!) = 145
Enfoque: la idea es iterar desde [1, N] y verificar si algún número entre el rango es un número fuerte o no . En caso afirmativo, imprima el número correspondiente; de lo contrario, verifique el siguiente número.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Store the factorial of all the
// digits from [0, 9]
int factorial[] = { 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 };
// Function to return true
// if number is strong or not
bool isStrong(int N)
{
// Converting N to String so that
// can easily access all it's digit
string num = to_string(N);
// sum will store summation of
// factorial of all digits
// of a number N
int sum = 0;
for(int i = 0; i < num.length(); i++)
{
sum += factorial[num[i] - '0'];
}
// Returns true of N is strong number
return sum == N;
}
// Function to print all
// strong number till N
void printStrongNumbers(int N)
{
// Iterating from 1 to N
for(int i = 1; i <= N; i++)
{
// Checking if a number is
// strong then print it
if (isStrong(i))
{
cout << i << " ";
}
}
}
// Driver Code
int main()
{
// Given number
int N = 1000;
// Function call
printStrongNumbers(N);
return 0;
}
// This code is contributed by rutvik_56
Java
// Java program for the above approach
class GFG {
// Store the factorial of all the
// digits from [0, 9]
static int[] factorial = { 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 };
// Function to return true
// if number is strong or not
public static boolean isStrong(int N)
{
// Converting N to String so that
// can easily access all it's digit
String num = Integer.toString(N);
// sum will store summation of
// factorial of all digits
// of a number N
int sum = 0;
for (int i = 0;
i < num.length(); i++) {
sum += factorial[Integer
.parseInt(num
.charAt(i)
+ "")];
}
// Returns true of N is strong number
return sum == N;
}
// Function to print all
// strong number till N
public static void
printStrongNumbers(int N)
{
// Iterating from 1 to N
for (int i = 1; i <= N; i++) {
// Checking if a number is
// strong then print it
if (isStrong(i)) {
System.out.print(i + " ");
}
}
}
// Driver Code
public static void
main(String[] args)
throws java.lang.Exception
{
// Given Number
int N = 1000;
// Function Call
printStrongNumbers(N);
}
}
Python3
# Python3 program for the
# above approach
# Store the factorial of
# all the digits from [0, 9]
factorial = [1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880]
# Function to return true
# if number is strong or not
def isStrong(N):
# Converting N to String
# so that can easily access
# all it's digit
num = str(N)
# sum will store summation
# of factorial of all
# digits of a number N
sum = 0
for i in range (len(num)):
sum += factorial[ord(num[i]) -
ord('0')]
# Returns true of N
# is strong number
if sum == N:
return True
else:
return False
# Function to print all
# strong number till N
def printStrongNumbers(N):
# Iterating from 1 to N
for i in range (1, N + 1):
# Checking if a number is
# strong then print it
if (isStrong(i)):
print (i, end = " ")
# Driver Code
if __name__ == "__main__":
# Given number
N = 1000
# Function call
printStrongNumbers(N)
# This code is contributed by Chitranayal
C#
// C# program for the above approach
using System;
class GFG{
// Store the factorial of all the
// digits from [0, 9]
static int[] factorial = { 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 };
// Function to return true
// if number is strong or not
public static bool isStrong(int N)
{
// Converting N to String so that
// can easily access all it's digit
String num = N.ToString();
// sum will store summation of
// factorial of all digits
// of a number N
int sum = 0;
for (int i = 0; i < num.Length; i++)
{
sum += factorial[int.Parse(num[i] + "")];
}
// Returns true of N is strong number
return sum == N;
}
// Function to print all
// strong number till N
public static void printStrongNumbers(int N)
{
// Iterating from 1 to N
for (int i = 1; i <= N; i++)
{
// Checking if a number is
// strong then print it
if (isStrong(i))
{
Console.Write(i + " ");
}
}
}
// Driver Code
public static void Main(String[] args)
{
// Given Number
int N = 1000;
// Function Call
printStrongNumbers(N);
}
}
// This code is contributed by sapnasingh4991
Javascript
<script>
// Javascript program for the above approach
// Store the factorial of all the
// digits from [0, 9]
let factorial = [ 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 ];
// Function to return true
// if number is strong or not
function isStrong(N)
{
// Converting N to String so that
// can easily access all it's digit
let num = N.toString();
// sum will store summation of
// factorial of all digits
// of a number N
let sum = 0;
for(let i = 0; i < num.length; i++)
{
sum += factorial[num[i] - '0'];
}
// Returns true of N is strong number
return sum == N;
}
// Function to print all
// strong number till N
function printStrongNumbers(N)
{
// Iterating from 1 to N
for(let i = 1; i <= N; i++)
{
// Checking if a number is
// strong then print it
if (isStrong(i))
{
document.write(i + " ");
}
}
}
// Driver Code
// Given number
let N = 1000;
// Function call
printStrongNumbers(N);
// This code is contributed by Mayank Tyagi
</script>
Producción:
1 2 145
Complejidad de tiempo: O(N)
Publicación traducida automáticamente
Artículo escrito por shubham prakash 1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA