打印移动方向，使您停留在[-k，+k]边界内
原文:https://www . geeksforgeeks . org/print-移动方向-这样-你就可以停留在-k-k-边界内/
给定一个由正整数 N 和整数 K 组成的数组 arr[] 。假设你从位置 0 开始，你可以从arr【0】开始通过a【I】位置向左或向右移动。任务是打印移动方向，通过向右或向左移动，您可以完成 N 步，而不超过 [-K，+K] 边界。如果不能执行步骤，打印 -1 。如果有多个答案，打印任意一个。 举例:
输入: arr[] = {40，50，60，40}，K = 120 输出: 右 右 左 右 说明: 由于 N = 4(数组中的元素个数) 我们需要从 arr[0]开始进行 4 次移动，使得 值不会超出[-120， 120] 移动 1:位置= 0 + 40 = 40 移动 2:位置= 40 + 50 = 90 移动 3:位置= 90–60 = 30 移动 4:位置= 30 + 50 = 80 输入: arr[] = {40，50，60，40}，K = 20 输出: -1
方法:可以按照以下步骤解决上述问题:
- 开始时将位置初始化为 0。
- 开始遍历所有的数组元素，
- 如果 a[i] +位置没有超过左右边界，那么移动将是“右”。
- 如果位置-a[I]没有超过左右边界，那么移动将是“左”。
- 如果在任何阶段两个条件都失败，则打印 -1 。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print steps such that
// they do not cross the boundary
void printSteps(int a[], int n, int k)
{
// To store the resultant string
string res = "";
// Initially at zero-th position
int position = 0;
int steps = 1;
// Iterate for every i-th move
for (int i = 0; i < n; i++) {
// Check for right move condition
if (position + a[i] <= k
&& position + a[i] >= (-k)) {
position += a[i];
res += "Right\n";
}
// Check for left move condition
else if (position - a[i] >= -k
&& position - a[i] <= k) {
position -= a[i];
res += "Left\n";
}
// No move is possible
else {
cout << -1;
return;
}
}
// Print the steps
cout << res;
}
// Driver code
int main()
{
int a[] = { 40, 50, 60, 40 };
int n = sizeof(a) / sizeof(a[0]);
int k = 120;
printSteps(a, n, k);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to print steps such that
// they do not cross the boundary
static void printSteps(int []a, int n, int k)
{
// To store the resultant string
String res = "";
// Initially at zero-th position
int position = 0;
//int steps = 1;
// Iterate for every i-th move
for (int i = 0; i < n; i++)
{
// Check for right move condition
if (position + a[i] <= k
&& position + a[i] >= (-k))
{
position += a[i];
res += "Right\n";
}
// Check for left move condition
else if (position - a[i] >= -k
&& position - a[i] <= k)
{
position -= a[i];
res += "Left\n";
}
// No move is possible
else
{
System.out.println(-1);
return;
}
}
// Print the steps
System.out.println(res);
}
// Driver code
public static void main (String[] args)
{
int []a = { 40, 50, 60, 40 };
int n = a.length;
int k = 120;
printSteps(a, n, k);
}
}
// This code is contributed by mits
Python 3
# Python3 implementation of the approach
# Function to print steps such that
# they do not cross the boundary
def printSteps(a, n, k):
# To store the resultant string
res = ""
# Initially at zero-th position
position = 0
steps = 1
# Iterate for every i-th move
for i in range(n):
# Check for right move condition
if (position + a[i] <= k and
position + a[i] >= -k):
position += a[i]
res += "Right\n"
# Check for left move condition
elif (position-a[i] >= -k and
position-a[i] <= k):
position -= a[i]
res += "Left\n"
# No move is possible
else:
print(-1)
return
print(res)
# Driver code
a = [40, 50, 60, 40]
n = len(a)
k = 120
printSteps(a, n, k)
# This code is contributed by Shrikant13
C
// C# implementation of the approach
using System;
class GFG
{
// Function to print steps such that
// they do not cross the boundary
static void printSteps(int []a, int n, int k)
{
// To store the resultant string
String res = "";
// Initially at zero-th position
int position = 0;
//int steps = 1;
// Iterate for every i-th move
for (int i = 0; i < n; i++)
{
// Check for right move condition
if (position + a[i] <= k
&& position + a[i] >= (-k))
{
position += a[i];
res += "Right\n";
}
// Check for left move condition
else if (position - a[i] >= -k
&& position - a[i] <= k)
{
position -= a[i];
res += "Left\n";
}
// No move is possible
else
{
Console.WriteLine(-1);
return;
}
}
// Print the steps
Console.Write(res);
}
// Driver code
static void Main()
{
int []a = { 40, 50, 60, 40 };
int n = a.Length;
int k = 120;
printSteps(a, n, k);
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to print steps such that
// they do not cross the boundary
function printSteps($a, $n, $k)
{
// To store the resultant string
$res = "";
// Initially at zero-th position
$position = 0;
$steps = 1;
// Iterate for every i-th move
for ($i = 0; $i < $n; $i++)
{
// Check for right move condition
if ($position + $a[$i] <= $k
&& $position + $a[$i] >= (-$k))
{
$position += $a[$i];
$res .= "Right\n";
}
// Check for left move condition
else if ($position - $a[$i] >= -$k &&
$position - $a[$i] <= $k)
{
$position -= $a[$i];
$res .= "Left\n";
}
// No move is possible
else
{
echo -1;
return;
}
}
// Print the steps
echo $res;
}
// Driver code
$a = array( 40, 50, 60, 40 );
$n = count($a);
$k = 120;
printSteps($a, $n, $k);
// This code is contributed by mits
?>
java 描述语言
<script>
// javascript implementation of the approach
// Function to print steps such that
// they do not cross the boundary
function printSteps(a , n , k) {
// To store the resultant string
var res = "";
// Initially at zero-th position
var position = 0;
// var steps = 1;
// Iterate for every i-th move
for (i = 0; i < n; i++) {
// Check for right move condition
if (position + a[i] <= k && position + a[i] >= (-k)) {
position += a[i];
res += "Right<br/>";
}
// Check for left move condition
else if (position - a[i] >= -k && position - a[i] <= k) {
position -= a[i];
res += "Left<br/>";
}
// No move is possible
else {
document.write(-1);
return;
}
}
// Print the steps
document.write(res);
}
// Driver code
var a = [ 40, 50, 60, 40 ];
var n = a.length;
var k = 120;
printSteps(a, n, k);
// This code is contributed by todaysgaurav
</script>
Output:
Right
Right
Left
Right
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