提交时间:2025-12-26 14:29:33

运行 ID: 77582

#include <iostream> using namespace std; int main() { int n, total, effective; cin >> n; cin >> total >> effective; double base = (double)effective / total; for(int i = 1; i < n; i++) { cin >> total >> effective; double rate = (double)effective / total; if(rate - base > 0.05) cout << "better" << endl; else if(base - rate > 0.05) cout << "worse" << endl; else cout << "same" << endl; } return 0; }