If there are "spikes" in two or more consecutive cells, then only the quantity of all previous coins needs to be counted; If there are no consecutive $2 $or more cells with spikes, it can be proven that we can definitely reach the last cell.

So this is the code: ~~~~~

include <bits/stdc++.h>

using namespace std;

int sum = 0; int main() { int t; cin >> t; while (t--) { int n, now = 0; string s; cin >> n >> s; int sum = 0; if(s.find("**") != string::npos) { //找到了连续的两根刺 for(int i = 0; i < s.find("**"); i++) { if(s[i] == '@') sum++; } cout << sum << endl; } else { for(int i = 0; i < s.length(); i++) { if(s[i] == '@') sum++; } cout << sum << endl; } } return 0; } ~~~~~