原文:
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
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）
= (-$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
output:
right
right
left
right
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