$O(nL\lg n)$
#include<iostream> #include<string> #include<tuple> #include<algorithm> using namespace std; int n; tuple<int, int, string> p[1000]; int main() { cin >> n; for (int i = 0; i < n; i++) { cin >> get<2>(p[i]); for (char c : get<2>(p[i])) if (c <= '9') get<1>(p[i]) += c - '0'; get<0>(p[i]) = get<2>(p[i]).size(); } sort(p, p + n); for (int i = 0; i < n; i++) cout << get<2>(p[i]) << endl; return 0; }
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