結果
問題 | No.968 引き算をして門松列(その3) |
ユーザー | hitonanode |
提出日時 | 2020-01-13 21:01:53 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 4,744 bytes |
コンパイル時間 | 1,411 ms |
コンパイル使用メモリ | 168,576 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-06-02 03:46:43 |
合計ジャッジ時間 | 2,469 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 1 ms
6,812 KB |
testcase_01 | AC | 1 ms
6,940 KB |
testcase_02 | AC | 2 ms
6,940 KB |
testcase_03 | WA | - |
testcase_04 | WA | - |
testcase_05 | WA | - |
testcase_06 | WA | - |
testcase_07 | WA | - |
testcase_08 | WA | - |
testcase_09 | WA | - |
testcase_10 | WA | - |
testcase_11 | WA | - |
ソースコード
#include <bits/stdc++.h> using namespace std; using lint = long long int; using pint = pair<int, int>; using plint = pair<lint, lint>; struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_; #define ALL(x) (x).begin(), (x).end() #define SZ(x) ((lint)(x).size()) #define POW2(n) (1LL << (n)) #define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++) #define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--) #define REP(i, n) FOR(i,0,n) #define IREP(i, n) IFOR(i,0,n) template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); } template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); } template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; } template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; } template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); } template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); } template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; } ///// This part below is only for debug, not used ///// template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; } template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; } template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; } template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; } template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; } template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; } template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; } template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; } template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; } #define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl; ///// END ///// /* #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #include <ext/pb_ds/tag_and_trait.hpp> using namespace __gnu_pbds; // find_by_order(), order_of_key() template<typename TK> using pbds_set = tree<TK, null_type, less<TK>, rb_tree_tag, tree_order_statistics_node_update>; template<typename TK, typename TV> using pbds_map = tree<TK, TV, less<TK>, rb_tree_tag, tree_order_statistics_node_update>; */ constexpr lint INF = 3e18; lint dfs(lint A, lint B, lint C, lint X, lint Y, lint Z, int depth) { if (A <= 0 or B <= 0 or C <= 0 or depth <= 0) return INF; if (A != B and B != C and C != A) return 0; lint ret = INF; mmin(ret, X + dfs(A - 1, B - 1, C, X, Y, Z, depth - 1)); mmin(ret, Y + dfs(A, B - 1, C - 1, X, Y, Z, depth - 1)); mmin(ret, Z + dfs(A - 1, B, C - 1, X, Y, Z, depth - 1)); return ret; } lint solve1(lint A, lint B, lint C, lint X, lint Y, lint Z) { // a < b > c にする lint sub = max({A - B, C - B, 0LL}); lint cost = sub * Z; A -= sub; C -= sub; return dfs(A, B, C, X, Y, Z, 6) + cost; } lint solve2(lint A, lint B, lint C, lint X, lint Y, lint Z) { // a > b < c にする lint cost = 0; lint suby = max(B - A + 1, 0LL); B -= suby, C -= suby; cost += suby * Y; lint subx = max(B - C + 1, 0LL); A -= subx, B -= subx; cost += subx * X; return dfs(A, B, C, X, Y, Z, 6) + cost; } int main() { int T; cin >> T; while (T--) { lint A, B, C, X, Y, Z; cin >> A >> B >> C >> X >> Y >> Z; lint ret = min(solve1(A, B, C, X, Y, Z), solve2(A, B, C, X, Y, Z)); cout << (ret < INF ? ret : -1) << endl; } }