Dados dos números positivos N y M , la tarea es comprobar si los pares de números dados (N, M) forman un Número de Compromiso o no.
Ejemplos:
Entrada: N = 48, M = 75
Salida: Sí
Explicación:
Los divisores propios de 48 son 1, 2, 3, 4, 6, 8, 12, 16, 24 La
suma de los divisores propios de 48 es 75( sum1 )
El los divisores de 75 son 1, 3, 5, 15, 25 La
suma de los divisores propios de 48 es 49( sum2 )
Dado que sum2 = N + 1, los pares dados forman números prometidos.
Entrada: N = 95, M = 55
Salida: No
Explicación:
Los divisores propios de 95 son 1, 5, 19 La
suma de los divisores propios de 48 es 25( sum1 )
Los divisores propios de 55 son 1, 5, 11 La
suma de los divisores propios divisores de 48 es 17( suma2 )
Dado que ni sum2 es igual a N + 1 ni sum1 es igual a M + 1, por lo tanto, los pares dados no forman números prometidos.
Acercarse:
- Encuentra la suma de los divisores propios de los números dados N y M .
- Si la suma de los divisores propios de N es igual a M + 1 o la suma de los divisores propios de M es igual a N + 1 , entonces los pares dados forman Números de Compromiso .
- De lo contrario, no forma un par de números prometidos.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check whether N is
// Perfect Square or not
bool isPerfectSquare(int N)
{
// Find sqrt
double sr = sqrt(N);
return (sr - floor(sr)) == 0;
}
// Function to check whether the given
// pairs of numbers is Betrothed Numbers
// or not
void BetrothedNumbers(int n, int m)
{
int Sum1 = 1;
int Sum2 = 1;
// For finding the sum of all the
// divisors of first number n
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
Sum1 += i
+ (isPerfectSquare(n)
? 0
: n / i);
}
}
// For finding the sum of all the
// divisors of second number m
for (int i = 2; i <= sqrt(m); i++) {
if (m % i == 0) {
Sum2 += i
+ (isPerfectSquare(m)
? 0
: m / i);
}
}
if ((n + 1 == Sum2)
&& (m + 1 == Sum1)) {
cout << "YES" << endl;
}
else {
cout << "NO" << endl;
}
}
// Driver Code
int main()
{
int N = 9504;
int M = 20734;
// Function Call
BetrothedNumbers(N, M);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to check whether N is
// Perfect Square or not
static boolean isPerfectSquare(int N)
{
// Find sqrt
double sr = Math.sqrt(N);
return (sr - Math.floor(sr)) == 0;
}
// Function to check whether the given
// pairs of numbers is Betrothed Numbers
// or not
static void BetrothedNumbers(int n, int m)
{
int Sum1 = 1;
int Sum2 = 1;
// For finding the sum of all the
// divisors of first number n
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
Sum1 += i
+ (isPerfectSquare(n)
? 0
: n / i);
}
}
// For finding the sum of all the
// divisors of second number m
for (int i = 2; i <= Math.sqrt(m); i++) {
if (m % i == 0) {
Sum2 += i
+ (isPerfectSquare(m)
? 0
: m / i);
}
}
if ((n + 1 == Sum2)
&& (m + 1 == Sum1)) {
System.out.print("YES" +"\n");
}
else {
System.out.print("NO" +"\n");
}
}
// Driver Code
public static void main(String[] args)
{
int N = 9504;
int M = 20734;
// Function Call
BetrothedNumbers(N, M);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program for the above approach
from math import sqrt,floor
# Function to check whether N is
# Perfect Square or not
def isPerfectSquare(N):
# Find sqrt
sr = sqrt(N)
return (sr - floor(sr)) == 0
# Function to check whether the given
# pairs of numbers is Betrothed Numbers
# or not
def BetrothedNumbers(n,m):
Sum1 = 1
Sum2 = 1
# For finding the sum of all the
# divisors of first number n
for i in range(2,int(sqrt(n))+1,1):
if (n % i == 0):
if (isPerfectSquare(n)):
Sum1 += i
else:
Sum1 += i + n/i
# For finding the sum of all the
# divisors of second number m
for i in range(2,int(sqrt(m))+1,1):
if (m % i == 0):
if (isPerfectSquare(m)):
Sum2 += i
else:
Sum2 += i + (m / i)
if ((n + 1 == Sum2) and (m + 1 == Sum1)):
print("YES")
else:
print("NO")
# Driver Code
if __name__ == '__main__':
N = 9504
M = 20734
# Function Call
BetrothedNumbers(N, M)
# This code is contributed by Surendra_Gangwar
C#
// C# program for the above approach
using System;
class GFG{
// Function to check whether N is
// perfect square or not
static bool isPerfectSquare(int N)
{
// Find sqrt
double sr = Math.Sqrt(N);
return (sr - Math.Floor(sr)) == 0;
}
// Function to check whether the given
// pairs of numbers is Betrothed numbers
// or not
static void BetrothedNumbers(int n, int m)
{
int Sum1 = 1;
int Sum2 = 1;
// For finding the sum of all the
// divisors of first number n
for(int i = 2; i <= Math.Sqrt(n); i++)
{
if (n % i == 0)
{
Sum1 += i + (isPerfectSquare(n) ?
0 : n / i);
}
}
// For finding the sum of all the
// divisors of second number m
for(int i = 2; i <= Math.Sqrt(m); i++)
{
if (m % i == 0)
{
Sum2 += i + (isPerfectSquare(m) ?
0 : m / i);
}
}
if ((n + 1 == Sum2) && (m + 1 == Sum1))
{
Console.Write("YES" + "\n");
}
else
{
Console.Write("NO" + "\n");
}
}
// Driver Code
public static void Main(String[] args)
{
int N = 9504;
int M = 20734;
// Function Call
BetrothedNumbers(N, M);
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript program for the above approach
// Function to check whether N is
// Perfect Square or not
function isPerfectSquare(N)
{
// Find sqrt
let sr = Math.sqrt(N);
return (sr - Math.floor(sr)) == 0;
}
// Function to check whether the given
// pairs of numbers is Betrothed Numbers
// or not
function BetrothedNumbers(n, m)
{
let Sum1 = 1;
let Sum2 = 1;
// For finding the sum of all the
// divisors of first number n
for (let i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
Sum1 += i
+ (isPerfectSquare(n)
? 0
: parseInt(n / i));
}
}
// For finding the sum of all the
// divisors of second number m
for (let i = 2; i <= Math.sqrt(m); i++) {
if (m % i == 0) {
Sum2 += i
+ (isPerfectSquare(m)
? 0
: parseInt(m / i));
}
}
if ((n + 1 == Sum2)
&& (m + 1 == Sum1)) {
document.write("YES");
}
else {
document.write("NO");
}
}
// Driver Code
let N = 9504;
let M = 20734;
// Function Call
BetrothedNumbers(N, M);
// This code is contributed by rishavmahato348.
</script>
Producción:
NO
Complejidad del Tiempo: O(√N + √M)
Publicación traducida automáticamente
Artículo escrito por hrishikeshkonderu y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA