在数轴中从零找到到达 X 的跳跃次数

给定一个整数 X。任务是找到从零开始到达数轴中点 X 的跳跃次数。
注意：第一次跳跃的长度可以是一个单位，每次连续跳跃的长度都将比前一次跳跃的长度长一个单位。每次跳跃都允许向左或向右移动。
例子：

Input : X = 8
Output : 4
Explanation : 
0 -> -1 -> 1 -> 4-> 8 are possible stages.

Input : X = 9
Output : 5
Explanation : 
0 -> -1 -> -3 -> 0 -> 4-> 9 are 
possible stages

方法：仔细观察，可以很容易地说：

如果您一直朝正确的方向跳跃，那么在跳跃n次后，您将到达p = 1 + 2 + 3 + 4 + … + n 点。
如果您在第 k 次跳跃时不是向右跳，而是向左跳，那么您将处于p – 2k点。
此外，通过仔细选择向左跳和向右跳，在 n 次跳跃后，您可以在 n * (n + 1) / 2 和 – (n * (n + 1) / 2) 之间的任何点与 n * (n + 1) / 2 相同的奇偶校验。

牢记以上几点，你要做的就是模拟跳跃的过程，总是向右跳跃，如果在某个点，你已经到达了一个与X具有相同奇偶校验的点，并且处于或超过X，你会有你的答案。
下面是上述方法的实现：

C++

// C++ program to find the number of jumps
// to reach X in the number line from zero
 
#include 
using namespace std;
 
// Utility function to calculate sum
// of numbers from 1 to x
int getsum(int x)
{
    return (x * (x + 1)) / 2;
}
 
// Function to find the number of jumps
// to reach X in the number line from zero
int countJumps(int n)
{
    // First make number positive
    // Answer will be same either it is
    // Positive or negative
    n = abs(n);
 
    // To store required answer
    int ans = 0;
 
    // Continue till number is lesser or not in same parity
    while (getsum(ans) < n or (getsum(ans) - n) & 1)
        ans++;
 
    // Return the required answer
    return ans;
}
 
// Driver code
int main()
{
    int n = 9;
 
    cout << countJumps(n);
 
    return 0;
}

Java

// Java program to find the number of jumps
// to reach X in the number line from zero
 
class GFG
{
     
// Utility function to calculate sum
// of numbers from 1 to x
static int getsum(int x)
{
    return (x * (x + 1)) / 2;
}
 
// Function to find the number of jumps
// to reach X in the number line from zero
static int countJumps(int n)
{
    // First make number positive
    // Answer will be same either it is
    // Positive or negative
    n = Math.abs(n);
 
    // To store required answer
    int ans = 0;
 
    // Continue till number is lesser
    // or not in same parity
    while (getsum(ans) < n ||
         ((getsum(ans) - n) & 1) > 0)
        ans++;
 
    // Return the required answer
    return ans;
}
 
// Driver code
public static void main(String args[])
{
    int n = 9;
 
    System.out.println(countJumps(n));
}
}
 
// This code is contributed by Ryuga

Python3

# Python 3 program to find the number of jumps
# to reach X in the number line from zero
 
# Utility function to calculate sum
# of numbers from 1 to x
def getsum(x):
    return int((x * (x + 1)) / 2)
 
# Function to find the number of jumps
# to reach X in the number line from zero
def countJumps(n):
     
    # First make number positive
    # Answer will be same either it is
    # Positive or negative
    n = abs(n)
 
    # To store the required answer
    ans = 0
 
    # Continue till number is lesser
    # or not in same parity
    while (getsum(ans) < n or
          (getsum(ans) - n) & 1):
        ans += 1
 
    # Return the required answer
    return ans
 
# Driver code
if __name__ == '__main__':
    n = 9
 
    print(countJumps(n))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find the number of jumps
// to reach X in the number line from zero
using System;
 
class GFG
{
     
// Utility function to calculate sum
// of numbers from 1 to x
static int getsum(int x)
{
    return (x * (x + 1)) / 2;
}
 
// Function to find the number of jumps
// to reach X in the number line from zero
static int countJumps(int n)
{
    // First make number positive
    // Answer will be same either it is
    // Positive or negative
    n = Math.Abs(n);
 
    // To store required answer
    int ans = 0;
 
    // Continue till number is lesser or not in same parity
    while (getsum(ans) < n || ((getsum(ans) - n) & 1)>0)
        ans++;
 
    // Return the required answer
    return ans;
}
 
// Driver code
static void Main()
{
    int n = 9;
 
    Console.WriteLine(countJumps(n));
}
}
 
// This code is contributed by mits

PHP

Javascript

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