使用之字形方式从(0，0)到达点(x，y)所需的步数
原文:https://www . geesforgeks . org/从 00 开始到达 xy 点所需的步数-使用之字形路线/
给定一个坐标 (x，y) 。任务是使用之字形方式计算从(0，0)到达点(x，y)所需的步数，并且不能直线行进超过 1 个单位。另外，开始沿着 Y 轴移动。 例如，我们可以通过下图所示的各种方式到达红色表示的点:
例:
Input: x = 4, y = 4
Output: 8
In the diagram above the line is passing
using 8 steps.
Input: x = 4, y = 3
Output: 9
Input: x = 2, y = 1
Output: 5
走近:通过画一个小图，我们可以看到两种情况:
- 情况 1 :如果 x 小于 y，那么答案永远是x+y+2 *(y-x)/2)。
- 情况 2 :如果 x 大于等于 y，那么答案永远是x+y+2 *(x-y)+1)/2)。
以下是上述方法的实现:
C++
// C++ program to find the number of steps
// required to reach (x, y) from (0, 0) following
// a zig-zag path
#include <bits/stdc++.h>
using namespace std;
// Function to return the required position
int countSteps(int x, int y)
{
if (x < y) {
return x + y + 2 * ((y - x) / 2);
}
else {
return x + y + 2 * (((x - y) + 1) / 2);
}
}
// Driver Code
int main()
{
int x = 4, y = 3;
cout << countSteps(x, y);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the number of steps
// required to reach (x, y) from (0, 0) following
// a zig-zag path
class GfG
{
// Function to return the required position
static int countSteps(int x, int y)
{
if (x < y)
{
return x + y + 2 * ((y - x) / 2);
}
else
{
return x + y + 2 * (((x - y) + 1) / 2);
}
}
// Driver Code
public static void main(String[] args)
{
int x = 4, y = 3;
System.out.println(countSteps(x, y));
}
}
// This code is contributed by Prerna Saini
Python 3
# Python3 program to find the number of
# steps required to reach (x, y) from
# (0, 0) following a zig-zag path
# Function to return the required position
def countSteps(x, y):
if x < y:
return x + y + 2 * ((y - x) // 2)
else:
return x + y + 2 * (((x - y) + 1) // 2)
# Driver Code
if __name__ == "__main__":
x, y = 4, 3
print(countSteps(x, y))
# This code is contributed by Rituraj Jain
C
// C# program to find the number of steps
// required to reach (x, y) from (0, 0)
// following a zig-zag path
using System;
class GfG
{
// Function to return the required position
static int countSteps(int x, int y)
{
if (x < y)
{
return x + y + 2 * ((y - x) / 2);
}
else
{
return x + y + 2 * (((x - y) + 1) / 2);
}
}
// Driver Code
public static void Main()
{
int x = 4, y = 3;
Console.WriteLine(countSteps(x, y));
}
}
// This code is contributed by Code_Mech.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the number of steps
// required to reach (x, y) from (0, 0)
// following a zig-zag path
// Function to return the required position
function countSteps($x, $y)
{
if ($x < $y)
{
return $x + $y + 2 *
(($y - $x) / 2);
}
else
{
return $x + $y + 2 *
((($x - $y) + 1) / 2);
}
}
// Driver Code
$x = 4; $y = 3;
echo(countSteps($x, $y));
// This code is contributed
// by Code_Mech.
?>
java 描述语言
<script>
// Javascript program to find the number of steps
// required to reach (x, y) from (0, 0) following
// a zig-zag path
// Function to return the required position
function countSteps(x, y)
{
if (x < y) {
return x + y + 2 * parseInt((y - x) / 2);
}
else {
return x + y + 2 * parseInt(((x - y) + 1) / 2);
}
}
// Driver Code
var x = 4, y = 3;
document.write(countSteps(x, y));
// This code is contributed by rrrtnx.
</script>
Output:
9
时间复杂度: O(1)
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