需要添加到 N 以获得 K 的前 M 个倍数的总和的值

给定三个正整数N 、 K和M ，任务是找到要与N相加的数以获得K的前M倍数之和。

例子：

Input: N = 17, K = 3, M = 4
Output: 13
Explanation:
Sum of first 4 multiples of 3 = (3 + 6 + 9 + 12) = 30.
Therefore, the value to be added to 17 is (30 – 17) = 13.
Therefore, the required output is 13.

Input: N = 5, K = 2, M = 1
Output: -3
Explanation:
Sum of first 1 multiples of 2 is 2.
The value to be added to 5 to get 2 is (2 – 5) = -3

处理方法：按照以下步骤解决问题：

1. 计算K的前M 个倍数的总和，这将等于K * (1 + 2 + 3 + … M) = K * M * (M + 1) / 2 。
2. 初始化一个变量，比如res ，以存储需要添加到N以获得sum 的数字。
3. 因此， res将等于sum – N 。打印res的值。

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 value
// to be added to N to obtain
// sum of first M multiples of K
static int printNumber(int N, int K, int M)
{
   
    // Store the sum of the
    // first M multiples of K
    int sum = K * (M * (M + 1) / 2);
     
    // Store the value to be
    // added to obtain N
    return sum - N;
}
 
// Driver Code
int main()
{
    // Input
    int N = 17;
    int K = 3;
    int M = 4;
    cout << printNumber(N, K, M);
    return 0;
}
 
// This code is contributed by shubhamsingh10

Java

// Java code of above approach
import java.util.*;
class GFG
{
 
  // Function to print the value
  // to be added to N to obtain
  // sum of first M multiples of K
  static int printNumber(int N, int K, int M)
  {
 
    // Store the sum of the
    // first M multiples of K
    int sum = K * (M * (M + 1) / 2);
 
    // Store the value to be
    // added to obtain N
    return sum - N;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    // Input
    int N = 17;
    int K = 3;
    int M = 4;
    System.out.print(printNumber(N, K, M));
  }
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function to print the value
# to be added to N to obtain
# sum of first M multiples of K
def printNumber(N, K, M):
 
    # Store the sum of the
    # first M multiples of K
    sum = K * (M * (M + 1) / 2)
 
    # Store the value to be
    # added to obtain N
    return sum - N
 
# Driver Code
 
# Input
N = 17
K = 3
M = 4
 
print(int(printNumber(N, K, M)))

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to print the value
  // to be added to N to obtain
  // sum of first M multiples of K
  static int printNumber(int N, int K, int M)
  {
 
    // Store the sum of the
    // first M multiples of K
    int sum = K * (M * (M + 1) / 2);
 
    // Store the value to be
    // added to obtain N
    return sum - N;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
 
    // Input
    int N = 17;
    int K = 3;
    int M = 4;
    Console.Write(printNumber(N, K, M));
  }
}
 
// This code is contributed by shubhamsingh10

Javascript

 
// JavaScript program for the above approach
 
// Function to print the value
// to be added to N to obtain
// sum of first M multiples of K
function printNumber(N, K, M)
{
     
    // Store the sum of the
    // first M multiples of K
    var sum = K * ((M * (M + 1)) / 2);
     
    // Store the value to be
    // added to obtain N
    return sum - N;
}
 
// Driver Code
 
// Input
var N = 17;
var K = 3;
var M = 4;
 
document.write(printNumber(N, K, M));
 
// This code is contributed by rdtank
 

输出

13

时间复杂度： O(1)
辅助空间： O(1)