找到范围 [L, R] 使得该范围内的数字之和等于 N

给定一个整数N (N ≠ 0)，任务是找到一个范围[L, R] (−10⁻¹⁸ < L < R < 10¹⁸)，使得该范围内所有整数的总和等于N。

L + (L+1) + … + (R−1) + R = N

例子：

Input : N = 3
Output: -2 3
Explanation: For L = -2 and R = -3 the sum becomes -2 + (-1) + 0 + 1 + 2 + 3 = 3

Input : N = -6
Output: -6 5
Explanation: The sum for this range [-6, 5] is -6 + (-5) + (-4) + (-3) + (-2) + (-1) + 0 + 1+ 2 + 3 + 4 + 5 = -6

编程需要懂一点英语

朴素方法：对于 L 的每个值，尝试找到满足条件 L + (L+1) + 的值 R。 . . + (R-1) + R = N，使用嵌套循环。
时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：由于 L 和 R 是整数并且也可以是负数，因此上述问题可以在 O(1) 中有效地解决。考虑以下观察：

对于 N 是一个正整数，我们可以考虑：

[−(N – 1)] + [−(N – 2)] + . . . -1 + 0 + 1 + . . . + (N − 1) + N =
-(N – 1) + (N – 1) – (N – 2) + (N – 2) + . . . + 1 – 1 + 0 + N = N
So, L = -(N – 1) and R = N

编程需要懂一点英语

同样，对于 N 为负数，我们可以考虑：

N + (N + 1) + . . . -1 + 0 + 1 + . . . + [-(N + 2)] + [-(N + 1)] =
(N + 1) – (N + 1) + (N + 2) – (N + 2) + . . . -1 + 1 + 0 + N = N
So L = N and R = -(N + 1)

编程需要懂一点英语

因此，这个问题在单位时间复杂度上的解是：

L = -(N – 1) and R = N, when N is a positive integer.

L = N and R = -(N + 1), when N is a negative integer.

编程需要懂一点英语

注意：这是满足问题要求的最长可能范围（即 R – L 具有最高值）。

下面是该方法的实现：

C++

// C++ code to implement above approach
 
#include 
using namespace std;
 
// Function to find two integers
void Find_Two_Intergers(long long int N)
{
    // Variable to store value of L and R
    long long int L, R;
 
    // When N is positive
    if (N > 0) {
        L = -(N - 1);
        R = N;
    }
 
    // When N is negative
    else {
        L = N;
        R = -(N + 1);
    }
    cout << L << " " << R;
}
 
// Driver Code
int main()
{
    long long int N = 3;
    Find_Two_Integers(N);
    return 0;
}

C

// C code to implement above approach
 
#include 
 
// Function to find two integers
void Find_Two_Intergers(long long int N)
{
    // Variable to store L and R
    long long int L, R;
 
    // When N is positive
    if (N > 0) {
        L = -(N - 1);
        R = N;
    }
 
    // When N is negative
    else {
        L = N;
        R = -(N + 1);
    }
    printf("%lld %lld", L, R);
}
 
// Driver code
int main()
{
    long long int N = 3;
    Find_Two_Integers(N);
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
 
class GFG
{
   
  // Function to find two integers
  static void Find_Two_Intergers(long  N)
  {
     
    // Variable to store value of L and R
    long  L, R;
 
    // When N is positive
    if (N > 0) {
      L = -(N - 1);
      R = N;
    }
 
    // When N is negative
    else {
      L = N;
      R = -(N + 1);
    }
    System.out.print( L + " " + R);
  }
 
  // Driver Code
  public static void main (String[] args) {
 
    long N = 3;
    Find_Two_Integers(N);
 
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python code to implement above approach
 
# Function to find two integers
def Find_Two_Intergers(N):
    # variable to store L and R
    L = 0
    R = 0
 
    # When N is positive
    if N > 0:
        L = -(N-1)
        R = N
 
    # When N is negative
    else:
        L = N
        R = -(N+1)
 
    print(L, R)
 
 
# Driver code
N = 3
Find_Two_Integers(N)

C#

// C# code for the above approach
using System;
 
class GFG
{
   
  // Function to find two integers
  static void Find_Two_Intergers(long  N)
  {
     
    // Variable to store value of L and R
    long  L, R;
 
    // When N is positive
    if (N > 0) {
      L = -(N - 1);
      R = N;
    }
 
    // When N is negative
    else {
      L = N;
      R = -(N + 1);
    }
    Console.Write( L + " " + R);
  }
 
  // Driver Code
  public static void Main (String[] args) {
 
    long N = 3;
    Find_Two_Integers(N);
 
  }
}
 
// This code is contributed by Saurabh Jaiswal

Javascript

输出

-2 3

时间复杂度： O(1)
辅助空间： O(1)