检查 N 在除以 K 以内的所有值时是否只留下不同的余数

给定两个整数N和K ，任务是检查N在除以[1, K]范围内的所有整数时是否只留下不同的余数。如果是，请打印Yes 。否则，打印No 。
例子：

Input: N = 5, K = 3 
Output: Yes 
Explanation: 
(5 % 1) == 0 
(5 % 2) == 1 
(5 % 3) == 2 
Since all the remainders {0, 1, 2} are distinct.
Input: N = 5, K = 4 
Output: No 
Explanation: 
(5 % 1) == 0 
(5 % 2) == 1 
(5 % 3) == 2 
(5 % 4) == 1, which is not distinct. 
 

方法：
请按照以下步骤解决问题：

• 初始化一个集合S
• 迭代范围[1, K] 。
• 在每次迭代中，检查N % i是否已经存在于Set S 中。
• 如果不存在，则将N % i插入集合S
• 否则，打印No并终止。

下面是上述方法的实现：

C++

// C++ Program to check if all
// remainders are distinct or not
#include 
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 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

// 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 s = new HashSet();
 
    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

# 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#

// 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 s = new HashSet();
 
    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

Javascript

 
// 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.

输出：

Yes

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