答案 `

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int money[50100];
int main(){
    int t;
    cin >>t;
    for (int i = 0;i<t;i++){
        int n;
        string l;
        cin >>n>>l;
        int wg = 0;
        int rmb = 0;
        for (int j = 0;;j++){
            if (wg >= n - 1){
                money[i] = rmb;
                break;
            }
            if(l[wg + 1] == '.'){
                if (wg + 1 >= n){
                    money[i] = rmb;
                    break;
                }
                wg++;
            }
            if(l[wg + 1] == '@'){
                if (wg + 1 >= n){
                    money[i] = rmb;
                    break;
                }
                wg++;
                rmb++;
            }
            else if(l[wg + 1] == '*' && wg + 2 >= l.size()){
                money[i] = rmb;
                break;
            }
            else if(l[wg + 1] == '*' && l[wg + 2] == '.'){
                if (wg + 2 >= n){
                    money[i] = rmb;
                    break;
                }
                wg += 2;
            }
            else if(l[wg + 1] == '*' && l[wg + 2] == '@'){
                if (wg + 2 >= n){
                    money[i] = rmb;
                    break;
                }
                wg += 2;
                rmb++;
            }
            else if(l[wg + 1] == '*' && l[wg + 2] == '*'){
                money[i] = rmb;
                break;
            }
        }
    }
    for (int i = 0;i<t;i++){
        cout <<money[i]<<endl;
    }
}

#include<bits/stdc++.h>
using namespace std;
int dfs(const string &a, int pos, vector<int> &ovo, int mxcoin = 0,    int step1 = 0, int step2 = 0) {
    if (pos >= a.size())
        return 0;
    if (ovo[pos] != -1)
        return ovo[pos];
    if (a[pos] == '@')
        mxcoin = 1;
    if (pos + 1 < a.size() && a[pos + 1] != '*') 
        step1 = dfs(a, pos + 1, ovo);
    if (pos + 2 < a.size() && a[pos + 2] != '*') 
        step2 = dfs(a, pos + 2, ovo);
    ovo[pos] = mxcoin + max(step1, step2);
    return ovo[pos];
}
int main() {
    int t = 0;
    cin >> t;
    while (t--) {
        int n = 0;
        cin >> n;
        string a;
        cin >> a;
        vector<int> ovo(n, -1);
        cout << dfs(a, 0, ovo) << endl;
    }
    return 0;
}

非常好 dfs ，爱来自基沃托斯 感觉这样写比楼上优雅，故发题解 还有，熟悉的，不使用 vector 只能度过一个比较失败的人生（bushi