找到范围【L，R】，使得该范围内的数字之和等于 N
原文:https://www . geesforgeks . org/find-the-range-l-r-so-sum-of-numbers-in-to-n/
给定一个整数 N (N ≠ 0)，任务是找到一个范围【l，r】(−10⁻⁸<l<r<10⁸)，使得该范围内所有整数之和等于 N 。
L+(L+1)+…+(R1)+R = N
示例:
输入: N = 3 输出: -2 3 解释:对于 L = -2 和 R = -3，和变为-2 + (-1) + 0 + 1 + 2 + 3 = 3
输入: N = -6 输出: -6 5 说明:这个范围的和[-6，5]是-6+(-5)+(-4)+(-3)+(-2)+(-1)+0+1+2+3+4+5 =-6
天真方法:对于 L 的每个值，试着找到一个满足 L + (L+1) +。。。+ (R-1) + R = N，使用嵌套循环。 时间复杂度: O(N 2 ) 辅助空间: O(1)
有效方法:由于 L 和 R 都是整数，也可以是负数，所以上述问题可以在 O(1)中有效地解决。考虑以下观察结果:
- 对于 N 为正整数我们可以考虑:
[(N–1)]+[(N–2)]+。。。-1 + 0 + 1 + .。。+(n1)+N = -(N–1)+(N–1)–(N–2)+(N–2)+。。。+1–1+0+N = N 所以，L =-(N–1)和 R = N
- 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 所以 L = N 和 R = -(N + 1)
因此，单位时间复杂度的这个问题的解决方案是:
当 N 为正整数时，l =-(N–1)和 R = N。
L = N，R = -(N + 1)，当 N 为负整数时。
注:这是满足问题要求的最大可能范围(即 R–L 值最高)。
以下是该方法的实施情况:
C++
// C++ code to implement above approach
#include <bits/stdc++.h>
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_Intergers(N);
return 0;
}
C
// C code to implement above approach
#include <stdio.h>
// 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_Intergers(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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_Intergers(N);
}
}
// This code is contributed by Potta Lokesh
Python 3
# 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_Intergers(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_Intergers(N);
}
}
// This code is contributed by Saurabh Jaiswal
java 描述语言
<script>
// Javascript code to implement above approach
// Function to find two integers
function Find_Two_Intergers(N) {
// Variable to store value of L and R
let L, R;
// When N is positive
if (N > 0) {
L = -(N - 1);
R = N;
}
// When N is negative
else {
L = N;
R = -(N + 1);
}
document.write(L + " " + R);
}
// Driver Code
let N = 3;
Find_Two_Intergers(N);
// This code is contributed by gfgking.
</script>
Output
-2 3
时间复杂度:O(1) T3】辅助空间: O(1)
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