方法思路

1. 我们需要模拟每一秒史莱姆的移动，然后统计没有史莱姆的格子数量
2. 对于每个格子，我们可以确定在第i秒时，哪些初始位置的史莱姆会移动到这个格子
3. 如果史莱姆向右移动(ai=1)，那么第i秒它会在位置i+1
4. 如果史莱姆向左移动(ai=0)，那么第i秒它会在位置i-1
5. 对于每一秒，我们计算每个格子是否有史莱姆，然后统计没有史莱姆的格子数量

代码实现

Java

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] directions = new int[n];
        
        for (int i = 0; i < n; i++) {
            directions[i] = sc.nextInt();
        }
        
        int[] result = countEmptyCells(n, directions);
        
        for (int i = 0; i < n; i++) {
            System.out.print(result[i] + (i == n - 1 ? "" : " "));
        }
    }
    
    public static int[] countEmptyCells(int n, int[] directions) {
        int[] result = new int[n];
        
        for (int second = 1; second <= n; second++) {
            boolean[] hasSlime = new boolean[n + 1]; // 1-indexed
            
            // 检查每个初始位置的史莱姆在当前秒的位置
            for (int initialPos = 1; initialPos <= n; initialPos++) {
                int currentPos;
                if (directions[initialPos - 1] == 0) { // 向左移动
                    currentPos = initialPos - second;
                } else { // 向右移动
                    currentPos = initialPos + second;
                }
                
                // 如果史莱姆还在地图上，标记该位置有史莱姆
                if (currentPos >= 1 && currentPos <= n) {
                    hasSlime[currentPos] = true;
                }
            }
            
            // 统计没有史莱姆的格子数量
            int emptyCells = 0;
            for (int pos = 1; pos <= n; pos++) {
                if (!hasSlime[pos]) {
                    emptyCells++;
                }
            }
            
            result[second - 1] = emptyCells;
        }
        
        return result;
    }
}

Python

def count_empty_cells(n, directions):
    result = []
    
    for second in range(1, n + 1):
        has_slime = [False] * (n + 1)  # 1-indexed
        
        # 检查每个初始位置的史莱姆在当前秒的位置
        for initial_pos in range(1, n + 1):
            if directions[initial_pos - 1] == 0:  # 向左移动
                current_pos = initial_pos - second
            else:  # 向右移动
                current_pos = initial_pos + second
            
            # 如果史莱姆还在地图上，标记该位置有史莱姆
            if 1 <= current_pos <= n:
                has_slime[current_pos] = True
        
        # 统计没有史莱姆的格子数量
        empty_cells = sum(1 for pos in range(1, n + 1) if not has_slime[pos])
        result.append(empty_cells)
    
    return result

n = int(input())
directions = list(map(int, input().split()))

result = count_empty_cells(n, directions)
print(" ".join(map(str, result)))

C++

#include <iostream>
#include <vector>
using namespace std;

vector<int> countEmptyCells(int n, vector<int>& directions) {
    vector<int> result(n);
    
    for (int second = 1; second <= n; second++) {
        vector<bool> hasSlime(n + 1, false); // 1-indexed
        
        // 检查每个初始位置的史莱姆在当前秒的位置
        for (int initialPos = 1; initialPos <= n; initialPos++) {
            int currentPos;
            if (directions[initialPos - 1] == 0) { // 向左移动
                currentPos = initialPos - second;
            } else { // 向右移动
                currentPos = initialPos + second;
            }
            
            // 如果史莱姆还在地图上，标记该位置有史莱姆
            if (currentPos >= 1 && currentPos <= n) {
                hasSlime[currentPos] = true;
            }
        }
        
        // 统计没有史莱姆的格子数量
        int emptyCells = 0;
        for (int pos = 1; pos <= n; pos++) {
            if (!hasSlime[pos]) {
                emptyCells++;
            }
        }
        
        result[second - 1] = emptyCells;
    }
    
    return result;
}

int main() {
    int n;
    cin >> n;
    
    vector<int> directions(n);
    for (int i = 0; i < n; i++) {
        cin >> directions[i];
    }
    
    vector<int> result = countEmptyCells(n, directions);
    
    for (int i = 0; i < n; i++) {
        cout << result[i];
        if (i < n - 1) cout << " ";
    }
    cout << endl;
    
    return 0;
}