左右交替走 N 步后的位置
给定三个整数 N、A 和 B，一个人站在 0 坐标，第一步 A 步向右移动，第二步 B 步向左移动，以此类推。任务是找出 N 步后他会在哪个坐标。 例:
输入: N = 3，A = 5，B = 2 输出: 8 向右 5，向左 2，向右 5，因此此人将在 8 结束。 输入: N = 5，A = 1，B = 10 输出: -17
逼近:由于人向右走奇数步，向左走偶数步，我们要找出两个方向的步数差。因此，得到的公式将是:
[((n+1)/2)*a - (n/2)*b]
以下是上述方法的实现:
C++
// C++ program to find the last coordinate
// where it ends his journey
#include <bits/stdc++.h>
using namespace std;
// Function to return the last destination
int lastCoordinate(int n, int a, int b)
{
return ((n + 1) / 2) * a - (n / 2) * b;
}
// Driver Code
int main()
{
int n = 3, a = 5, b = 2;
cout << lastCoordinate(n, a, b);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the last coordinate
// where it ends his journey
import java.util.*;
class solution
{
// Function to return the last destination
static int lastCoordinate(int n, int a, int b)
{
return ((n + 1) / 2) * a - (n / 2) * b;
}
// Driver Code
public static void main(String args[])
{
int n = 3, a = 5, b = 2;
System.out.println(lastCoordinate(n, a, b));
}
}
计算机编程语言
# Python3 program to find the
# last coordinate where it
# ends his journey
# Function to return the
# last destination
def lastCoordinate(n, a, b):
return (((n + 1) // 2) *
a - (n // 2) * b)
# Driver Code
n = 3
a = 5
b = 2
print(lastCoordinate(n, a, b))
# This code is contributed
# by Sanjit_Prasad
C
// C# program to find the last coordinate
// where it ends his journey
using System;
class GFG
{
// Function to return the last destination
public static int lastCoordinate(int n,
int a, int b)
{
return ((n + 1) / 2) * a - (n / 2) * b;
}
// Driver Code
public static void Main(string[] args)
{
int n = 3, a = 5, b = 2;
Console.WriteLine(lastCoordinate(n, a, b));
}
}
// This code is contributed by Shrikant13
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the last coordinate
// where it ends his journey
// Function to return the last destination
function lastCoordinate($n, $a, $b)
{
return (($n + 1) / 2) * $a -
(int)($n / 2) * $b;
}
// Driver Code
$n = 3; $a = 5; $b = 2;
echo lastCoordinate($n, $a, $b);
// This code is contributed by inder_verma..
?>
java 描述语言
<script>
// Javascript program to find
// the last coordinate
// where it ends his journey
// Function to return the last destination
function lastCoordinate(n, a, b)
{
return (parseInt(n + 1) / 2) * a -
parseInt(n / 2) * b;
}
// Driver Code
let n = 3, a = 5, b = 2;
document.write(lastCoordinate(n, a, b));
</script>
Output:
8
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