Dados dos números X e Y , la tarea es encontrar la probabilidad de que un divisor positivo arbitrario de 10 X sea un múltiplo entero de 10 Y.
Nota: Y debe ser <= X.
Ejemplos:
Entrada: X = 2, Y = 1
Salida: 4/9
Explicación:
Los divisores positivos de 10 2 son 1, 2, 4, 5, 10, 20, 25, 50, 100. Un total de 9.
Múltiplos de 10 1 hasta 10 2 son 10, 20, 50, 100. Un total de 4.
P(divisor de 10 2 es múltiplo de 10 1 ) = 4/9.
Entrada: X = 99, Y = 88
Salida: 9/625
Explicación:
A = 10 99 , B = 10 88
P(divisor de 10 99 es múltiplo de 10 88 ) = 9/625
Prerrequisitos: Número total de divisores de un número
Enfoque:
Para resolver el problema, necesitamos observar los siguientes pasos:
- Todos los divisores de 10 X serán de la forma:
(2 P * 5 Q ), donde 0 <= P, Q <= X
- Encuentra el número de factores primos de 10 X
10 X = 2 X * 5 X
- Por lo tanto, el número total de divisores de 10 X será: ( X + 1 ) * ( X + 1 ) .
- Ahora, considera todos los múltiplos de 10 Y que pueden ser divisores potenciales de 10 X . También son de la forma:
( 2 A * 5 B ), donde Y <= A, B <= X.
- Entonces, el conteo de divisores potenciales de 10 X que también son múltiplos de 10 Y es ( X – Y + 1 ) * ( X – Y + 1 ) .
- Por lo tanto, la probabilidad requerida es (( X – Y + 1 ) * ( X – Y + 1 )) / (( X + 1 ) * ( X + 1 )) . Calcular el valor de la expresión para valores dados de X e Y nos da la probabilidad requerida.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10
#include <bits/stdc++.h>
using namespace std;
#define int long long int
// Function to calculate and print
// the required probability
void prob(int x, int y)
{
// Count of potential divisors
// of X-th power of 10 which are
// also multiples of Y-th power
// of 10
int num = abs(x - y + 1)
* abs(x - y + 1);
// Count of divisors of X-th
// power of 10
int den = (x + 1) * (x + 1);
// Calculate GCD
int gcd = __gcd(num, den);
// Print the reduced
// fraction probability
cout << num / gcd << "/"
<< den / gcd << endl;
}
// Driver Code
signed main()
{
int X = 2, Y = 1;
prob(X, Y);
return 0;
}
Java
// Java program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10
import java.util.*;
class GFG{
// Function to calculate and print
// the required probability
static void prob(int x, int y)
{
// Count of potential divisors
// of X-th power of 10 which are
// also multiples of Y-th power
// of 10
int num = Math.abs(x - y + 1) *
Math.abs(x - y + 1);
// Count of divisors of X-th
// power of 10
int den = (x + 1) * (x + 1);
// Calculate GCD
int gcd = __gcd(num, den);
// Print the reduced
// fraction probability
System.out.print(num / gcd + "/" +
den / gcd + "\n");
}
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// Driver code
public static void main(String[] args)
{
int X = 2, Y = 1;
prob(X, Y);
}
}
// This code is contributed by amal kumar choubey
Python3
# Python3 program to find the probability
# of an arbitrary positive divisor of
# Xth power of 10 to be a multiple of
# Yth power of 10
from math import *
# Function to calculate and print
# the required probability
def prob(x, y):
# Count of potential divisors
# of X-th power of 10 which are
# also multiples of Y-th power
# of 10
num = abs(x - y + 1) * abs(x - y + 1)
# Count of divisors of X-th
# power of 10
den = (x + 1) * (x + 1)
# Calculate GCD
gcd1 = gcd(num, den)
# Print the reduced
# fraction probability
print(num // gcd1, end = "")
print("/", end = "")
print(den // gcd1)
# Driver Code
if __name__ == '__main__':
X = 2
Y = 1
prob(X, Y)
# This code is contributed by Surendra_Gangwar
C#
// C# program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10
using System;
class GFG{
// Function to calculate and print
// the required probability
static void prob(int x, int y)
{
// Count of potential divisors
// of X-th power of 10 which are
// also multiples of Y-th power
// of 10
int num = Math.Abs(x - y + 1) *
Math.Abs(x - y + 1);
// Count of divisors of X-th
// power of 10
int den = (x + 1) * (x + 1);
// Calculate GCD
int gcd = __gcd(num, den);
// Print the reduced
// fraction probability
Console.Write(num / gcd + "/" +
den / gcd + "\n");
}
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// Driver code
public static void Main(string[] args)
{
int X = 2, Y = 1;
prob(X, Y);
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// JavaScript program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10
// Function to calculate and print
// the required probability
function prob(x, y)
{
// Count of potential divisors
// of X-th power of 10 which are
// also multiples of Y-th power
// of 10
var num = Math.abs(x - y + 1) *
Math.abs(x - y + 1);
// Count of divisors of X-th
// power of 10
var den = (x + 1) * (x + 1);
// Calculate GCD
var gcd = __gcd(num, den);
// Print the reduced
// fraction probability
document.write(num / gcd + "/" +
den / gcd + "\n");
}
function __gcd(a, b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// Driver Code
var X = 2, Y = 1;
prob(X, Y);
// This code is contributed by Khushboogoyal499
</script>
Producción:
4/9
Complejidad del tiempo: O(log(N))
Publicación traducida automáticamente
Artículo escrito por greenindia y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA