给定一个N x M大小的网格和初始位置(X, Y)以及一系列K 次移动。
每个移动包含 2 个整数Dx和Dy 。
单次移动包括以下操作，这些操作也可以执行尽可能多的次数：
X = X + Dx and Y = Y + Dy.
任务是计算执行的移动次数。
请注意，在每个时间点，当前位置都必须在网格内。
例子：
Input: N = 4, M = 5, X = 1, Y = 1, k[] = {{1, 1}, {1, 1}, {0, -2}}
Output: 4
The values of (X, Y) are as follows:
Steps in Move 1: (1, 1) -> (2, 2) -> (3, 3) -> (4, 4) i.e. 3 moves
Move 2 is not possible as we cannot move to (5, 5) which is out of the bounds of the grid
Steps in Move 3: (4, 4) -> (4, 2) i.e. a single move and (4, 0) will be out of the bound so no further moves possible.
Total moves = 3 + 1 = 4
Input: N = 10, M = 10, X = 3, Y = 3, k[] = {{1, 1}, {-1, -1}}
Output: 16
方法：可以观察到，我们可以独立处理X和Y ，并将它们合并为当前移动中的步数=最小值（ X的步数， Y的步数）。要找到X的移动次数，我们可以使用Dx的符号来知道我们正在接近哪一端并找到到那一端的距离，速度为Dx ，我们可以找出它需要的步数。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
#define int long long
using namespace std;
// Function to return the count of
// possible steps in a single direction
int steps(int cur, int x, int n)
{
// It can cover infinite steps
if (x == 0)
return INT_MAX;
// We are approaching towards X = N
if (x > 0)
return abs((n - cur) / x);
// We are approaching towards X = 1
else
return abs((cur - 1) / x);
}
// Function to return the count of steps
int countSteps(int curx, int cury, int n, int m,
vector > moves)
{
int count = 0;
int k = moves.size();
for (int i = 0; i < k; i++) {
int x = moves[i].first;
int y = moves[i].second;
// Take the minimum of both moves independently
int stepct
= min(steps(curx, x, n), steps(cury, y, m));
// Update count and current positions
count += stepct;
curx += stepct * x;
cury += stepct * y;
}
return count;
}
// Driver code
main()
{
int n = 4, m = 5, x = 1, y = 1;
vector > moves = { { 1, 1 },
{ 1, 1 },
{ 0, -2 } };
int k = moves.size();
cout << countSteps(x, y, n, m, moves);
return 0;
} Java
// Java implementation of the above approach
class GFG
{
// Function to return the count of
// possible steps in a single direction
static int steps(int cur, int x, int n)
{
// It can cover infinite steps
if (x == 0)
return Integer.MAX_VALUE;
// We are approaching towards X = N
if (x > 0)
return Math.abs((n - cur) / x);
// We are approaching towards X = 1
else
return Math.abs((cur - 1) / x);
}
// Function to return the count of steps
static int countSteps(int curx, int cury,
int n, int m,
int[][] moves)
{
int count = 0;
int k = moves.length;
for (int i = 0; i < k; i++)
{
int x = moves[i][0];
int y = moves[i][1];
// Take the minimum of
// both moves independently
int stepct = Math.min(steps(curx, x, n),
steps(cury, y, m));
// Update count and current positions
count += stepct;
curx += stepct * x;
cury += stepct * y;
}
return count;
}
// Driver code
public static void main(String[] args)
{
int n = 4, m = 5, x = 1, y = 1;
int[][] moves = { { 1, 1 }, { 1, 1 },
{ 0, -2 } };
System.out.print(countSteps(x, y, n, m, moves));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 implementation of the approach
# Function to return the count of
# possible steps in a single direction
def steps(cur, x, n):
# It can cover infinite steps
if x == 0:
return float('inf')
# We are approaching towards X = N
elif x > 0:
return abs((n - cur) // x)
# We are approaching towards X = 1
else:
return abs(int((cur - 1) / x))
# Function to return the count of steps
def countSteps(curx, cury, n, m, moves):
count = 0
k = len(moves)
for i in range(0, k):
x = moves[i][0]
y = moves[i][1]
# Take the minimum of both moves
# independently
stepct = min(steps(curx, x, n),
steps(cury, y, m))
# Update count and current positions
count += stepct
curx += stepct * x
cury += stepct * y
return count
# Driver code
if __name__ == "__main__":
n, m, x, y = 4, 5, 1, 1
moves = [[1, 1], [1, 1], [0, -2]]
print(countSteps(x, y, n, m, moves))
# This code is contributed
# by Rituraj JainC#
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the count of
// possible steps in a single direction
static int steps(int cur, int x, int n)
{
// It can cover infinite steps
if (x == 0)
return int.MaxValue;
// We are approaching towards X = N
if (x > 0)
return Math.Abs((n - cur) / x);
// We are approaching towards X = 1
else
return Math.Abs((cur - 1) / x);
}
// Function to return the count of steps
static int countSteps(int curx, int cury,
int n, int m,
int[,] moves)
{
int count = 0;
int k = moves.GetLength(0);
for (int i = 0; i < k; i++)
{
int x = moves[i, 0];
int y = moves[i, 1];
// Take the minimum of
// both moves independently
int stepct = Math.Min(steps(curx, x, n),
steps(cury, y, m));
// Update count and current positions
count += stepct;
curx += stepct * x;
cury += stepct * y;
}
return count;
}
// Driver code
public static void Main(String[] args)
{
int n = 4, m = 5, x = 1, y = 1;
int[,] moves = { { 1, 1 }, { 1, 1 },
{ 0, -2 } };
Console.Write(countSteps(x, y, n, m, moves));
}
}
// This code is contributed by Rajput-JiPHP
0)
return floor(abs(($n - $cur) / $x));
// We are approaching towards X = 1
else
return floor(abs(($cur - 1) / $x));
}
// Function to return the count of steps
function countSteps($curx, $cury, $n, $m, $moves)
{
$count = 0;
$k = sizeof($moves);
for ($i = 0; $i < $k; $i++)
{
$x = $moves[$i][0];
$y = $moves[$i][1];
// Take the minimum of both moves independently
$stepct = min(steps($curx, $x, $n), steps($cury, $y, $m));
// Update count and current positions
$count += $stepct;
$curx += $stepct * $x;
$cury += $stepct * $y;
}
return $count;
}
// Driver code
$n = 4 ;
$m = 5 ;
$x = 1 ;
$y = 1 ;
$moves = array( array( 1, 1 ),
array( 1, 1 ),
array( 0, -2 ) );
$k = sizeof($moves);
echo countSteps($x, $y, $n, $m, $moves);
// This code is contributed by Ryuga
?>Javascript
输出：
4时间复杂度： O(N)