最多N的最大数，其与X的模数等于X的Y模

给定三个正整数X ， Y和N ，使得Y ，任务是从范围[0，N]中找到最大数，该范围的模数与X等于Y模X。

例子：

Input: X = 10, Y = 5, N = 15
Output: 15
Explanation:
The value of 15 % 10 (= 5) and 5 % 10 (= 5) are equal.
Therefore, the required output is 15.

Input: X = 5, Y = 0, N = 4
Output: 0

方法：可以根据以下观察结果解决给定问题：

• 由于Y小于X ，则Y％X必须为Y。因此，从具有X的模量为Y的范围[0，N]中找到最大值的想法。
• 假定最大数量，例如num = N ，以X为Y得到余数模。
• 用N％X的余数减去N以得到0的余数，然后将Y加。然后，带有X的那个数字的余数将是Y。
• 检查数字是否小于N。如果发现为真，则设置num =(N – N％X + Y) 。
• 否则，再次用X的值减去数字，即num =(N – N％X –(X – Y)) ，以从区间[0，N]中获得最大值。
• 数学上：
  • 如果(N – N％X + Y)≤N ，则设置num =(N – N％X + Y) 。
  • 否则，更新num =(N – N％X –(X – Y)) 。

请按照以下步骤解决问题：

• 初始化一个变量，例如num ，以存储具有[ Y，X ]范围内的余数Y％X的最大值。
• 如果(N – N％X + Y)≤N ，则更新num =(N – N％X + Y) 。
• 否则，更新num =(N – N％X –(X – Y)) 。
• 完成上述步骤后，打印num的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to print the largest
// number upto N whose modulus
// with X is same as Y % X
long long maximumNum(long long X,
                     long long Y,
                     long long N)
{
    // Stores the required number
    long long num = 0;
 
    // Update num as the result
    if (N - N % X + Y <= N) {
 
        num = N - N % X + Y;
    }
    else {
        num = N - N % X - (X - Y);
    }
 
    // Return the resultant number
    return num;
}
 
// Driver Code
int main()
{
    long long X = 10;
    long long Y = 5;
    long long N = 15;
 
    cout << maximumNum(X, Y, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG
{
 
  // Function to print the largest
  // number upto N whose modulus
  // with X is same as Y % X
  static long maximumNum(long X, long Y, long N)
  {
     
    // Stores the required number
    long num = 0;
 
    // Update num as the result
    if (N - N % X + Y <= N)
    {
      num = N - N % X + Y;
    }
    else
    {
      num = N - N % X - (X - Y);
    }
 
    // Return the resultant number
    return num;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    long X = 10;
    long Y = 5;
    long N = 15;
 
    System.out.println(maximumNum(X, Y, N));
  }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function to print the largest
# number upto N whose modulus
# with X is same as Y % X
def maximumNum(X, Y, N):
   
    # Stores the required number
    num = 0
 
    # Update num as the result
    if (N - N % X + Y <= N):
        num = N - N % X + Y
    else:
        num = N - N % X - (X - Y)
 
    # Return the resultant number
    return num
 
# Driver Code
if __name__ == '__main__':
    X = 10
    Y = 5
    N = 15
 
    print (maximumNum(X, Y, N))
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
class GFG {
 
  // Function to print the largest
  // number upto N whose modulus
  // with X is same as Y % X
  static long maximumNum(long X, long Y, long N)
  {
 
    // Stores the required number
    long num = 0;
 
    // Update num as the result
    if (N - N % X + Y <= N) {
      num = N - N % X + Y;
    }
    else {
      num = N - N % X - (X - Y);
    }
 
    // Return the resultant number
    return num;
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    long X = 10;
    long Y = 5;
    long N = 15;
 
    Console.WriteLine(maximumNum(X, Y, N));
  }
}
 
// This code is contributed by ukasp.

输出：

15

时间复杂度： O(1)
辅助空间： O(1)