Compruebe si N deja solo residuos distintos en la división por todos los valores hasta K

Dados dos números enteros N y K , la tarea es verificar si N deja solo residuos distintos cuando se divide por todos los números enteros en el rango [1, K] . Si es así, imprima Sí . De lo contrario , imprima No.
Ejemplos: 
 

Entrada: N = 5, K = 3 
Salida: Sí 
Explicación: 
(5 % 1) == 0 
(5 % 2) == 1 
(5 % 3) == 2 
Dado que todos los residuos {0, 1, 2} son distinto.
Entrada: N = 5, K = 4 
Salida: No 
Explicación: 
(5 % 1) == 0 
(5 % 2) == 1 
(5 % 3) == 2 
(5 % 4) == 1, que no es distinto. 
 

Enfoque: 
siga los pasos que se indican a continuación para resolver el problema: 
 

• Inicializar un conjunto S
• Iterar sobre el rango [1, K] .
• En cada iteración, verifique si N % i ya está presente en el Conjunto S o no.
• Si no está presente, inserte N % i en el conjunto S
• De lo contrario, imprima No y termine.

A continuación se muestra la implementación del enfoque anterior:
 

// C++ Program to check if all
// remainders are distinct or not
#include <bits/stdc++.h>
using namespace std;
 
// Function to check and return
// if all remainders are distinct
bool is_distinct(long long n, long long k)
{
 
    // Stores the remainder
    unordered_set<long long> s;
 
    for (int i = 1; i <= k; i++) {
 
        // Calculate the remainder
        long long tmp = n % i;
 
        // If remainder already occurred
        if (s.find(tmp) != s.end()) {
            return false;
        }
 
        // Insert into the set
        s.insert(tmp);
    }
 
    return true;
}
 
// Driver Code
int main()
{
 
    long long N = 5, K = 3;
 
    if (is_distinct(N, K))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

// Java program to check if all
// remainders are distinct or not
import java.util.*;
 
class GFG{
 
// Function to check and return
// if all remainders are distinct
static boolean is_distinct(long n, long k)
{
 
    // Stores the remainder
    HashSet<Long> s = new HashSet<Long>();
 
    for(int i = 1; i <= k; i++)
    {
         
        // Calculate the remainder
        long tmp = n % i;
 
        // If remainder already occurred
        if (s.contains(tmp))
        {
            return false;
        }
 
        // Insert into the set
        s.add(tmp);
    }
    return true;
}
 
// Driver Code
public static void main(String[] args)
{
    long N = 5, K = 3;
 
    if (is_distinct(N, K))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by gauravrajput1

# Python3 program to check if all
# remainders are distinct or not
 
# Function to check and return
# if all remainders are distinct
def is_distinct(n, k):
 
    # Stores the remainder
    s = set()
 
    for i in range(1, k + 1):
 
        # Calculate the remainder
        tmp = n % i
 
        # If remainder already occurred
        if (tmp in s):
            return False
 
        # Insert into the set
        s.add(tmp)
 
    return True
 
# Driver Code
if __name__ == '__main__':
 
    N = 5
    K = 3
     
    if (is_distinct(N, K)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by Shivam Singh

// C# program to check if all
// remainders are distinct or not
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check and return
// if all remainders are distinct
static bool is_distinct(long n, long k)
{
 
    // Stores the remainder
    HashSet<long> s = new HashSet<long>();
 
    for(int i = 1; i <= k; i++)
    {
         
        // Calculate the remainder
        long tmp = n % i;
 
        // If remainder already occurred
        if (s.Contains(tmp))
        {
            return false;
        }
 
        // Insert into the set
        s.Add(tmp);
    }
    return true;
}
 
// Driver Code
public static void Main(String[] args)
{
    long N = 5, K = 3;
 
    if (is_distinct(N, K))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by gauravrajput1

<script>
 
// Javascript program to check if all
// remainders are distinct or not
  
// Function to check and return
// if all remainders are distinct
function is_distinct(n, k)
{
   
    // Stores the remainder
    let s = new Set();
   
    for(let i = 1; i <= k; i++)
    {
           
        // Calculate the remainder
        let tmp = n % i;
   
        // If remainder already occurred
        if (s.has(tmp))
        {
            return false;
        }
   
        // Insert leto the set
        s.add(tmp);
    }
    return true;
}
 
// Driver code
 
    let N = 5, K = 3;
   
    if (is_distinct(N, K))
        document.write("Yes");
    else
        document.write("No");
 
// This code is contributed by souravghosh0416.
</script>

Yes

Complejidad temporal: O(K) 
Espacio auxiliar: O(K)