k天后的活动和非活动细胞
给定一个大小为 n 的二进制数组，其中n > 3 。数组中的真（或 1）值表示活动，假（或 0）表示不活动。给定一个数字 k，任务是找出 k 天后活动和非活动单元的数量。每天之后，如果左右单元格不相同，则第 i 个单元格的状态变为活动状态；如果左右单元格相同（均为 0 或均为 1），则第 i 个单元格的状态变为不活动状态。
由于在最左侧单元格之前和最右侧单元格之后没有单元格，因此最左侧单元格之前和最右侧单元格之后的值单元格始终被视为 0（或非活动）。
例子：
Input : cells[] = {1, 0, 1, 1}, k = 2
Output : Active cells = 3, Inactive cells = 1
After 1 day, cells[] = {0, 0, 1, 1}
After 2 days, cells[] = {0, 1, 1, 1}
Input : cells[] = {0, 1, 0, 1, 0, 1, 0, 1}, k = 3
Output: Active Cells = 2 , Inactive Cells = 6
Explanation :
After 1 day, cells[] = {1, 0, 0, 0, 0, 0, 0, 0}
After 2 days, cells[] = {0, 1, 0, 0, 0, 0, 0, 0}
After 3 days, cells[] = {1, 0, 1, 0, 0, 0, 0, 0}
Input : cells[] = {0, 1, 1, 1, 0, 1, 1, 0}, k = 4
Output: Active Cells = 3 , Inactive Cells = 5唯一重要的是确保我们维护给定数组的副本，因为我们需要在第二天更新以前的值。下面是详细步骤。
- 首先，我们将 cells[] 数组复制到 temp[] 数组中，并根据给定条件对 temp[] 数组进行更改。
- 在该条件下，如果第 i 个单元格的直接左右单元格不活跃或第二天活跃，则 i 变为不活跃，即； (cells[i-1] == 0 and cells[i+1] == 0) 或者 (cells[i-1] == 1 and cells[i+1] == 1) then cells[i] = 0 ，这些条件可以使用单元格[i-1] 和单元格[i+1] 的异或来应用。
- 对于第 0 个索引单元 temp[0] = 0^cells[1] 和第 (n-1)'th 个索引单元 temp[n-1] = 0^cells[n-2]。
- 现在对于索引 1 到 n-2，执行以下操作 temp[i] = cells[i-1] ^ cells[i+1]
- 重复这个过程，直到 k 天结束。
以下是上述步骤的实现。
C++
// C++ program to count active and inactive cells after k
// days
#include
using namespace std;
// cells[] - store current status of cells
// n - Number of cells
// temp[] - to perform intermediate operations
// k - number of days
// active - count of active cells after k days
// inactive - count of active cells after k days
void activeAndInactive(bool cells[], int n, int k)
{
// copy cells[] array into temp [] array
bool temp[n];
for (int i=0; i Java
// Java program to count active and
// inactive cells after k days
class GFG {
// cells[] - store current status
// of cells n - Number of cells
// temp[] - to perform intermediate operations
// k - number of days
// active - count of active cells after k days
// inactive - count of active cells after k days
static void activeAndInactive(boolean cells[],
int n, int k)
{
// copy cells[] array into temp [] array
boolean temp[] = new boolean[n];
for (int i = 0; i < n; i++)
temp[i] = cells[i];
// Iterate for k days
while (k-- > 0) {
// Finding next values for corner cells
temp[0] = false ^ cells[1];
temp[n - 1] = false ^ cells[n - 2];
// Compute values of intermediate cells
// If both cells active or inactive, then
// temp[i]=0 else temp[i] = 1.
for (int i = 1; i <= n - 2; i++)
temp[i] = cells[i - 1] ^ cells[i + 1];
// Copy temp[] to cells[] for next iteration
for (int i = 0; i < n; i++)
cells[i] = temp[i];
}
// count active and inactive cells
int active = 0, inactive = 0;
for (int i = 0; i < n; i++)
if (cells[i] == true)
active++;
else
inactive++;
System.out.print("Active Cells = " + active + ", " +
"Inactive Cells = " + inactive);
}
// Driver code
public static void main(String[] args)
{
boolean cells[] = {false, true, false, true,
false, true, false, true};
int k = 3;
int n = cells.length;
activeAndInactive(cells, n, k);
}
}
// This code is contributed by Anant Agarwal.Python3
# Python program to count
# active and inactive cells after k
# days
# cells[] - store current
# status of cells
# n - Number of cells
# temp[] - to perform
# intermediate operations
# k - number of days
# active - count of active
# cells after k days
# inactive - count of active
# cells after k days
def activeAndInactive(cells,n,k):
# copy cells[] array into temp [] array
temp=[]
for i in range(n+1):
temp.append(False)
for i in range(n):
temp[i] = cells[i]
# Iterate for k days
while (k >0):
# Finding next values for corner cells
temp[0] = False^cells[1]
temp[n-1] = False^cells[n-2]
# Compute values of intermediate cells
# If both cells active or
# inactive, then temp[i]=0
# else temp[i] = 1.
for i in range(1,n-2+1):
temp[i] = cells[i-1] ^ cells[i+1]
# Copy temp[] to cells[]
# for next iteration
for i in range(n):
cells[i] = temp[i]
k-=1
# count active and inactive cells
active = 0
inactive = 0;
for i in range(n):
if(cells[i] == True):
active+=1
else:
inactive+=1
print("Active Cells =",active," , "
, "Inactive Cells =",
inactive)
# Driver code
cells = [False, True, False, True,
False, True, False, True]
k = 3
n =len(cells)
activeAndInactive(cells, n, k)
# This code is contributed
# by Anant Agarwal.C#
// C# program to count active and
// inactive cells after k days
using System;
class GFG {
// cells[] - store current status
// of cells n - Number of cells
// temp[] - to perform intermediate
// operations k - number of days
// active - count of active cells
// after k days inactive - count
// of active cells after k days
static void activeAndInactive(bool []cells,
int n, int k)
{
// copy cells[] array into
// temp [] array
bool []temp = new bool[n];
for (int i = 0; i < n; i++)
temp[i] = cells[i];
// Iterate for k days
while (k-- > 0) {
// Finding next values
// for corner cells
temp[0] = false ^ cells[1];
temp[n - 1] = false ^ cells[n - 2];
// Compute values of intermediate cells
// If both cells active or inactive, then
// temp[i]=0 else temp[i] = 1.
for (int i = 1; i <= n - 2; i++)
temp[i] = cells[i - 1] ^ cells[i + 1];
// Copy temp[] to cells[]
// for next iteration
for (int i = 0; i < n; i++)
cells[i] = temp[i];
}
// count active and inactive cells
int active = 0, inactive = 0;
for (int i = 0; i < n; i++)
if (cells[i] == true)
active++;
else
inactive++;
Console.Write("Active Cells = " + active + ", " +
"Inactive Cells = " + inactive);
}
// Driver code
public static void Main()
{
bool []cells = {false, true, false, true,
false, true, false, true};
int k = 3;
int n = cells.Length;
activeAndInactive(cells, n, k);
}
}
// This code is contributed by Nitin Mittal.PHP
Javascript
输出：
Active Cells = 2 , Inactive Cells = 6