solution

• 好运数：去除任意位末尾数位 所得到的数都满足能够被当前数位整除

#include<iostream>
#include<string>
using namespace std;
int main(){int k, flag;string s;cin >> k;while(k--){flag = 1;cin >> s;for(int i = 1; i <= s.size() && flag; i++){if(atoi(s.substr(0, i).c_str()) % i) flag = 0;}if(flag) cout << "Yes" << endl;else cout << "No" << endl;}return 0;
}