(i * j)％的最小可能值2019

给定两个整数L和R ，任务是找到(i * j)％2019的最小可能值，其中L≤i

例子：

Input: L = 2020, R = 2040
Output: 2
(2020*2021)%2019 = 2

Input: L = 3, R = 4
Output: 12

方法：

• 如果R – L≥2019，则答案为0，因为我们将得到一个可以被2019整除的数字，该数字为0。
• 如果R – L <2019，则我们可以运行嵌套循环并找到最小值。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
const int MOD = 2019;
  
// Function to return the minimum
// possible value of (i * j) % 2019
int min_modulo(int l, int r)
{
    // If we can get a number
    // divisible by 2019
    if (r - l >= MOD)
        return 0;
    else {
  
        // Find the minimum value
        // by running nested loops
        int ans = MOD - 1;
        for (int i = l; i <= r; i++) {
            for (int j = i + 1; j <= r; j++) {
                ans = min(ans, (i * j) % MOD);
            }
        }
        return ans;
    }
}
  
// Driver code
int main()
{
    int l = 2020, r = 2040;
  
    cout << min_modulo(l, r);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
static int MOD = 2019;
  
// Function to return the minimum
// possible value of (i * j) % 2019
static int min_modulo(int l, int r)
{
    // If we can get a number
    // divisible by 2019
    if (r - l >= MOD)
        return 0;
    else 
    {
  
        // Find the minimum value
        // by running nested loops
        int ans = MOD - 1;
        for (int i = l; i <= r; i++) 
        {
            for (int j = i + 1; j <= r; j++)
            {
                ans = Math.min(ans, (i * j) % MOD);
            }
        }
        return ans;
    }
}
  
// Driver code
public static void main(String []args) 
{
    int l = 2020, r = 2040;
  
    System.out.println(min_modulo(l, r));
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach 
MOD = 2019; 
  
# Function to return the minimum 
# possible value of (i * j) % 2019 
def min_modulo(l, r) :
  
    # If we can get a number 
    # divisible by 2019 
    if (r - l >= MOD) :
        return 0; 
    else :
  
        # Find the minimum value 
        # by running nested loops 
        ans = MOD - 1; 
        for i in range(l, r + 1) :
            for j in range(i + 1, r + 1) :
                ans = min(ans, (i * j) % MOD); 
          
        return ans; 
  
# Driver code 
if __name__ == "__main__" :
      
    l = 2020; r = 2040;
      
    print(min_modulo(l, r)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
  
class GFG 
{
static int MOD = 2019;
  
// Function to return the minimum
// possible value of (i * j) % 2019
static int min_modulo(int l, int r)
{
    // If we can get a number
    // divisible by 2019
    if (r - l >= MOD)
        return 0;
    else
    {
  
        // Find the minimum value
        // by running nested loops
        int ans = MOD - 1;
        for (int i = l; i <= r; i++) 
        {
            for (int j = i + 1; j <= r; j++)
            {
                ans = Math.Min(ans, (i * j) % MOD);
            }
        }
        return ans;
    }
}
  
// Driver code
public static void Main(String []args) 
{
    int l = 2020, r = 2040;
  
    Console.WriteLine(min_modulo(l, r));
}
}
  
// This code is contributed by Rajput-Ji

输出：

2