原文:
给定二维矩阵和整数“k”，任务是在“k”次迭代后预测矩阵，如下所示:
- element 1 in the current matrix will remain 1 in the next iteration only if it is surrounded by multiple 1s, where 0 < = range 1a < = a < = range 1b.
- the element 0 in the current matrix will only become 1 in the next iteration if it is surrounded by the number of b of 1, where 0<= range 0a<= b<= range 0b.
我们用一个例子来理解一下:
约束: 1<= k<= 100000 0<=范围 1a、范围 1b、范围 0a、范围 0b < = 8
- 在上述单元格(0，0)的图像中，单元格在第一次迭代中为“0”，但由于它仅被一个包含“1”的相邻单元格包围，因此不在范围[range0a，range0b]内。因此它将继续保持“0”。
- 对于第二次迭代，单元格(0，0)为 0，但这次它被两个包含“1”的单元格包围，两个单元格位于范围[range0a，range0b]内。因此，它在下一次(第二次)迭代中变为“1”。
示例:
输入:范围 1a = 2 范围 1b = 2 范围 0a = 2 范围 0b = 3 k = 1 输出: 0 1 1 0 0 1 1 1 0 0 1 0 0 1 0 输入:范围 1a = 2 范围 1b = 2 范围 0a = 2 范围 0b = 3
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 语言(一种计算机语言，尤用于创建网站)
// 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 chitranayal
c
// 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 29ajaykumar
服务器端编程语言（professional hypertext preprocessor 的缩写）
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
java 描述语言
output:
0 1 0 0
0 1 0 0
1 1 1 0
0 0 1 0
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