这是我的任务：
有一个由8个细胞排列成直线的集落，每天每个细胞与其相邻的细胞（邻居）竞争。每天，对于每个小区，如果其邻居都是活动的或者都是非活动的，则该小区在第二天变为不活动。否则它会在第二天活跃起来。
假设：两端的两个小区具有单个相邻小区，因此可以假设另一个相邻小区始终处于非活动状态。即使在更新了单元状态之后。考虑其更新其他单元状态的过去状态。同时更新所有单元的单元信息。
写一个函数cellCompete，它接受一个8元素的整数单元格数组，代表8个单元格的当前状态，一个整数天代表模拟的天数。整数值1表示活动单元格，值0表示非活动单元格。
程序：
int* cellCompete(int* cells,int days) { //write your code here } //function signature ends Test Case 1: INPUT: [1,0,0,0,0,1,0,0],1 EXPECTED RETURN VALUE: [0,1,0,0,1,0,1,0] Test Case 2: INPUT: [1,1,1,0,1,1,1,1,],2 EXPECTED RETURN VALUE: [0,0,0,0,0,1,1,0]
这是上面针对该问题给出的问题陈述。我为这个问题编写的代码如下。但输出与输入相同。
#include<iostream>
using namespace std;
// signature function to solve the problem
int *cells(int *cells,int days)
{ int previous=0;
for(int i=0;i<days;i++)
{
if(i==0)
{
if(cells[i+1]==0)
{
previous=cells[i];
cells[i]=0;
}
else
{
cells[i]=0;
}
if(i==days-1)
{
if(cells[days-2]==0)
{
previous=cells[days-1];
cells[days-1]=0;
}
else
{
cells[days-1]=1;
}
}
if(previous==cells[i+1])
{
previous=cells[i];
cells[i]=0;
}
else
{
previous=cells[i];
cells[i]=1;
}
}
}
return cells;
}
int main()
{
int array[]={1,0,0,0,0,1,0,0};
int *result=cells(array,8);
for(int i=0;i<8;i++)
cout<<result[i];
}
我无法得到错误，我认为我的逻辑错了。我们可以在这里应用动态编程吗？如果我们可以如何？
您的程序不区分模拟天数和单元格数。
#include <stdio.h>
int main() {
int days,ind,arr[8],outer;
for(ind=0;ind<8;scanf("%d ",&arr[ind]),ind++); //Reading the array
scanf("%d",&days);
int dupArr[8];
for(outer=0;outer<days;outer++){ //Number of days to simulate
for(ind=0;ind<8;ind++){ //Traverse the whole array
//cells on the ends have single adjacent cell, so the other adjacent cell can be assumsed to be always inactive
if(ind==0){
if(arr[ind+1]==0)
dupArr[ind]=0;
else
dupArr[ind]=1;
}
else if(ind==7){
if(arr[ind-1]==0)
dupArr[ind]=0;
else
dupArr[ind]=1;
}
else{
if((arr[ind-1]==0&&arr[ind+1]==0) || (arr[ind-1]==1&&arr[ind+1]==1)){// if its neighbours are both active or both inactive, the cell becomes inactive the next day
dupArr[ind]=0;
}
else //otherwise it becomes active the next day
dupArr[ind]=1;
}
}
for(ind=0;ind<8;ind++){
arr[ind]=dupArr[ind]; //Copying the altered array to original array, so that we can alter it n number of times.
}
}
for(ind=0;ind<8;ind++)
printf("%d ",arr[ind]);//Displaying output
return 0;
}
这是我几个月前创建的代码，
func competeCell(cell []uint, days uint) []uint{
n := len(cell)
temp := make([]uint, n)
for i :=0; i < n; i ++ {
temp[i] = cell[i]
}
for days > 0 {
temp[0] = 0 ^ cell[1]
temp[n-1] = 0 ^ cell[n-2]
for i := 1; i < n-2 +1; i++ {
temp[i] = cell[i-1] ^ cell[i +1]
}
for i:=0; i < n; i++ {
cell[i] = temp[i]
}
days -= 1
}
return cell
}
#include <bits/stdc++.h>
using namespace std;
int* cellCompete(int* cells,int days)
{
for(int j=0; j<days; j++)
{
int copy_cells[10];
for(int i=1; i<9; i++)
copy_cells[i]=cells[i-1];
copy_cells[0]=0;copy_cells[9]=0;
for(int i=0; i<8; i++)
cells[i]=copy_cells[i]==copy_cells[i+2]?0:1;
}
return cells;
}
int main()
{
int arr[8]={1,1,1,0,1,1,1,1};
int arr2[8]={1,0,0,0,0,1,0,0};
cellCompete(arr2,1);
for(int i=0; i<8; i++)
{
cout<<arr2[i]<<" ";
}
}
private List<Integer> finalStates = new ArrayList<>();
public static void main(String[] args) {
// int arr[] = { 1, 0, 0, 0, 0, 1, 0, 0 };
// int days = 1;
EightHousePuzzle eightHousePuzzle = new EightHousePuzzle();
int arr[] = { 1, 1, 1, 0, 1, 1, 1, 1 };
int days = 2;
eightHousePuzzle.cellCompete(arr, days);
}
public List<Integer> cellCompete(int[] states, int days) {
List<Integer> currentCellStates = Arrays.stream(states).boxed().collect(Collectors.toList());
return getCellStateAfterNDays(currentCellStates, days);
}
private List<Integer> getCellStateAfterNDays(List<Integer> currentCellStates, int days) {
List<Integer> changedCellStates = new ArrayList<>();
int stateUnoccupied = 0;
if (days != 0) {
for (int i1 = 0; i1 < currentCellStates.size(); i1++) {
if (i1 == 0) {
changedCellStates.add(calculateCellState(stateUnoccupied, currentCellStates.get(i1 + 1)));
} else if (i1 == 7) {
changedCellStates.add(calculateCellState(currentCellStates.get(i1 - 1), stateUnoccupied));
} else {
changedCellStates
.add(calculateCellState(currentCellStates.get(i1 - 1), currentCellStates.get(i1 + 1)));
}
}
if (days == 1) {
System.out.println("days ==1 hit");
finalStates = new ArrayList<>(changedCellStates);
return finalStates;
}
days = days - 1;
System.out.println("Starting recurssion");
getCellStateAfterNDays(changedCellStates, days);
}
return finalStates;
}
private int calculateCellState(int previousLeft, int previousRight) {
if ((previousLeft == 0 && previousRight == 0) || (previousLeft == 1 && previousRight == 1)) {
// the state gets now changed to 0
return 0;
}
// the state gets now changed to 0
return 1;
}
这是一些甜蜜的小python代码：
def cell(arr, days):
new = arr[:] #get a copy of the array
n = len(arr)
if n == 1: print [0] #when only 1 node, return [0]
for _ in range(days):
new[0] = arr[1] #determine the edge nodes first
new[n - 1] = arr[n - 2]
for i in range(1, n-1):
new[i] = 1 - (arr[i-1] == arr[i+1]) #logic for the rest nodes
arr = new[:] #update the list for the next day
return new
arr = [1, 1, 1, 0, 1, 1, 1, 1]
days = 2
print cell(arr, days)
public class Colony {
public static int[] cellCompete(int[] cell, int day) {
int[] ar = new int[10];
for(int i=1; i<9; i++) {
ar[i] = cell[i-1];
}
while(day-- >0) {
int temp = 0;
for(int i=1; i<9; i++) {
if(Math.abs(temp-ar[i+1])==1) {
temp = ar[i];
ar[i] = 1;
}
else {
temp = ar[i];
ar[i] = 0;
}
}
}
return ar;
}
public static void main(String[] args) {
int[] cell = {1,0,1,1,0,1,0,1};
int day = 1;
cell = cellCompete(cell, day);
for(int i=1; i<9; i++) {
System.out.print(cell[i]+" ");
}
}
}