给定一个二维矩阵和一个整数“ K”，任务是在“ K”次迭代后预测矩阵，如下所示：
- An element
1in current matrix remains1in the next iteration only if it is surrounded byAnumber of1s; where0 <= range1a <= A <= range1b. - An element
0in current matrix becomes1in the next iteration only if it is surrounded byBnumber of1s; where0 <= range0a <= B <= range0b.
让我们通过一个例子来理解这一点：
Constraints:
1 <= K <= 100000
0 <= range1a, range1b, range0a, range0b <= 8
- 在上面的单元格(0，0)的图像中，单元格在第一次迭代中为’0’，但由于它仅被一个相邻的包含’1’的单元格包围，而该单元格不在[range0a，range0b]范围内。因此它将继续保持为“ 0”。
- 对于第二次迭代，单元格(0，0)为0，但这一次它被两个包含’1’的单元格包围，并且两个落差的范围为[range0a，range0b]，因此在下一次(第二次)迭代中它变为’1′ 。
例子：
Input: range1a = 2
range1b = 2
range0a = 2
range0b = 3
K = 1
Output:
0 1 1 0
0 1 1 1
1 0 0 1
0 0 1 0
Input: range1a = 2
range1b = 2
range0a = 2
range0b = 3
K = 2
Output:
1 0 0 1
1 0 0 0
0 0 0 0
0 1 0 1
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Dimension of Array
#define N 4
void predictMatrix(int arr[N][N],
int range1a,
int range1b,
int range0a,
int range0b,
int K,
int b[N][N])
{
// Count of 1s
int c = 0;
while (K--) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
c = 0;
// Counting all neighbouring 1s
if (i > 0 && arr[i - 1][j] == 1)
c++;
if (j > 0 && arr[i][j - 1] == 1)
c++;
if (i > 0 && j > 0
&& arr[i - 1][j - 1] == 1)
c++;
if (i < N - 1 && arr[i + 1][j] == 1)
c++;
if (j < N - 1 && arr[i][j + 1] == 1)
c++;
if (i < N - 1 && j < N - 1
&& arr[i + 1][j + 1] == 1)
c++;
if (i < N - 1 && j > 0
&& arr[i + 1][j - 1] == 1)
c++;
if (i > 0 && j < N - 1
&& arr[i - 1][j + 1] == 1)
c++;
// Comparing the number of
// neighbouring 1s with
// given ranges
if (arr[i][j] == 1) {
if (c >= range1a && c <= range1b)
b[i][j] = 1;
else
b[i][j] = 0;
}
if (arr[i][j] == 0) {
if (c >= range0a && c <= range0b)
b[i][j] = 1;
else
b[i][j] = 0;
}
}
}
// Copying changes to
// the main matrix
for (int k = 0; k < N; k++)
for (int m = 0; m < N; m++)
arr[k][m] = b[k][m];
}
}
// Driver code
int main()
{
int arr[N][N] = { 0, 0, 0, 0,
0, 1, 1, 0,
0, 0, 1, 0,
0, 1, 0, 1 };
int range1a = 2, range1b = 2;
int range0a = 2, range0b = 3;
int K = 3, b[N][N] = { 0 };
// Function call to calculate
// the resultant matrix
// after 'K' iterations.
predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b);
// Printing Result
for (int i = 0; i < N; i++) {
cout << endl;
for (int j = 0; j < N; j++)
cout << b[i][j] << " ";
}
return 0;
} Java
// Java implementation of the approach
public class GFG{
// Dimension of Array
final static int N = 4 ;
static void predictMatrix(int arr[][],
int range1a,
int range1b,
int range0a,
int range0b,
int K,
int b[][])
{
// Count of 1s
int c = 0;
while (K != 0) {
K--;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
c = 0;
// Counting all neighbouring 1s
if (i > 0 && arr[i - 1][j] == 1)
c++;
if (j > 0 && arr[i][j - 1] == 1)
c++;
if (i > 0 && j > 0
&& arr[i - 1][j - 1] == 1)
c++;
if (i < N - 1 && arr[i + 1][j] == 1)
c++;
if (j < N - 1 && arr[i][j + 1] == 1)
c++;
if (i < N - 1 && j < N - 1
&& arr[i + 1][j + 1] == 1)
c++;
if (i < N - 1 && j > 0
&& arr[i + 1][j - 1] == 1)
c++;
if (i > 0 && j < N - 1
&& arr[i - 1][j + 1] == 1)
c++;
// Comparing the number of
// neighbouring 1s with
// given ranges
if (arr[i][j] == 1) {
if (c >= range1a && c <= range1b)
b[i][j] = 1;
else
b[i][j] = 0;
}
if (arr[i][j] == 0) {
if (c >= range0a && c <= range0b)
b[i][j] = 1;
else
b[i][j] = 0;
}
}
}
// Copying changes to
// the main matrix
for (int k = 0; k < N; k++)
for (int m = 0; m < N; m++)
arr[k][m] = b[k][m];
}
}
// Driver code
public static void main(String []args)
{
int arr[][] = { {0, 0, 0, 0},
{0, 1, 1, 0},
{0, 0, 1, 0},
{0, 1, 0, 1 } };
int range1a = 2, range1b = 2;
int range0a = 2, range0b = 3;
int K = 3;
int b[][] = new int[N][N] ;
// Function call to calculate
// the resultant matrix
// after 'K' iterations.
predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b);
// Printing Result
for (int i = 0; i < N; i++) {
System.out.println();
for (int j = 0; j < N; j++)
System.out.print(b[i][j]+ " ");
}
}
// This Code is contributed by Ryuga
}Python 3
# Python3 implementation of the approach
# Dimension of Array
N = 4
def predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b):
# Count of 1s
c = 0
while (K):
for i in range(N) :
for j in range(N):
c = 0
# Counting all neighbouring 1s
if (i > 0 and arr[i - 1][j] == 1):
c += 1
if (j > 0 and arr[i][j - 1] == 1):
c += 1
if (i > 0 and j > 0 and
arr[i - 1][j - 1] == 1):
c += 1
if (i < N - 1 and arr[i + 1][j] == 1):
c += 1
if (j < N - 1 and arr[i][j + 1] == 1):
c += 1
if (i < N - 1 and j < N - 1
and arr[i + 1][j + 1] == 1):
c += 1
if (i < N - 1 and j > 0
and arr[i + 1][j - 1] == 1):
c += 1
if (i > 0 and j < N - 1
and arr[i - 1][j + 1] == 1):
c += 1
# Comparing the number of neighbouring
# 1s with given ranges
if (arr[i][j] == 1) :
if (c >= range1a and c <= range1b):
b[i][j] = 1
else:
b[i][j] = 0
if (arr[i][j] == 0):
if (c >= range0a and c <= range0b):
b[i][j] = 1
else:
b[i][j] = 0
K -= 1
# Copying changes to the main matrix
for k in range(N):
for m in range( N):
arr[k][m] = b[k][m]
# Driver code
if __name__ == "__main__":
arr = [[0, 0, 0, 0],
[0, 1, 1, 0],
[0, 0, 1, 0],
[0, 1, 0, 1]]
range1a = 2
range1b = 2
range0a = 2
range0b = 3
K = 3
b = [[0 for x in range(N)]
for y in range(N)]
# Function call to calculate
# the resultant matrix
# after 'K' iterations.
predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b)
# Printing Result
for i in range( N):
print()
for j in range(N):
print(b[i][j], end = " ")
# This code is contributed
# by ChitraNayalC#
// C# implementation of the approach
using System;
class GFG
{
// Dimension of Array
readonly static int N = 4 ;
static void predictMatrix(int [,]arr, int range1a,
int range1b, int range0a,
int range0b, int K, int [,]b)
{
// Count of 1s
int c = 0;
while (K != 0)
{
K--;
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
c = 0;
// Counting all neighbouring 1s
if (i > 0 && arr[i - 1, j] == 1)
c++;
if (j > 0 && arr[i, j - 1] == 1)
c++;
if (i > 0 && j > 0
&& arr[i - 1, j - 1] == 1)
c++;
if (i < N - 1 && arr[i + 1, j] == 1)
c++;
if (j < N - 1 && arr[i, j + 1] == 1)
c++;
if (i < N - 1 && j < N - 1 &&
arr[i + 1, j + 1] == 1)
c++;
if (i < N - 1 && j > 0 &&
arr[i + 1, j - 1] == 1)
c++;
if (i > 0 && j < N - 1 &&
arr[i - 1, j + 1] == 1)
c++;
// Comparing the number of
// neighbouring 1s with
// given ranges
if (arr[i,j] == 1)
{
if (c >= range1a && c <= range1b)
b[i, j] = 1;
else
b[i, j] = 0;
}
if (arr[i,j] == 0)
{
if (c >= range0a && c <= range0b)
b[i, j] = 1;
else
b[i, j] = 0;
}
}
}
// Copying changes to the main matrix
for (int k = 0; k < N; k++)
for (int m = 0; m < N; m++)
arr[k, m] = b[k, m];
}
}
// Driver code
public static void Main()
{
int [,]arr = { {0, 0, 0, 0},
{0, 1, 1, 0},
{0, 0, 1, 0},
{0, 1, 0, 1 } };
int range1a = 2, range1b = 2;
int range0a = 2, range0b = 3;
int K = 3;
int [,]b = new int[N, N];
// Function call to calculate
// the resultant matrix
// after 'K' iterations.
predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b);
// Printing Result
for (int i = 0; i < N; i++)
{
Console.WriteLine();
for (int j = 0; j < N; j++)
Console.Write(b[i, j] + " ");
}
}
}
// This code is contributed by 29AjayKumarPHP
0 && $arr[$i - 1][$j] == 1)
$c++;
if ($j > 0 && $arr[$i][$j - 1] == 1)
$c++;
if ($i > 0 && $j > 0 && $arr[$i - 1][$j - 1] == 1)
$c++;
if ($i < $N - 1 && $arr[$i + 1][$j] == 1)
$c++;
if ($j < $N - 1 && $arr[$i][$j + 1] == 1)
$c++;
if ($i < $N - 1 && $j < $N - 1 &&
$arr[$i + 1][$j + 1] == 1)
$c++;
if ($i < $N - 1 && $j > 0 && $arr[$i + 1][$j - 1] == 1)
$c++;
if ($i > 0 && $j < $N - 1 && $arr[$i - 1][$j + 1] == 1)
$c++;
// Comparing the number of
// neighbouring 1s with
// given ranges
if ($arr[$i][$j] == 1)
{
if ($c >= $range1a && $c <= $range1b)
$b[$i][$j] = 1;
else
$b[$i][$j] = 0;
}
if ($arr[$i][$j] == 0) {
if ($c >= $range0a && $c <= $range0b)
$b[$i][$j] = 1;
else
$b[$i][$j] = 0;
}
}
}
// Copying changes to
// the main matrix
for ($k = 0; $k < $N; $k++)
for ($m = 0; $m < $N; $m++)
$arr[$k][$m] = $b[$k][$m];
}
return $b;
}
// Driver code
$N = 4;
$arr= array(array(0, 0, 0, 0),
array(0, 1, 1, 0),
array(0, 0, 1, 0),
array(0, 1, 0, 1));
$range1a = 2; $range1b = 2;
$range0a = 2; $range0b = 3;
$K = 3; $b = array(array(0));
// Function call to calculate
// the resultant matrix
// after 'K' iterations.
$b1 = predictMatrix($arr, $range1a, $range1b,
$range0a, $range0b, $K, $b);
// Printing Result
for ($i = 0; $i < $N; $i++)
{
echo "\n";
for ($j = 0; $j < $N; $j++)
echo $b1[$i][$j] . " ";
}
// This code is contributed by Akanksha Rai输出：
0 1 0 0
0 1 0 0
1 1 1 0
0 0 1 0