結果

問題 No.968 引き算をして門松列（その３）
ユーザー hitonanodehitonanode
提出日時 2020-01-13 21:05:47
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 4,740 bytes
コンパイル時間 1,690 ms
コンパイル使用メモリ 168,504 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-06-02 03:54:49
合計ジャッジ時間 2,049 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
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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 2 ms
6,816 KB
testcase_01 AC 1 ms
6,940 KB
testcase_02 WA -
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 -
権限があれば一括ダウンロードができます

ソースコード

diff # 

#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 subz = max({A - B, C - B, 0LL});
    lint cost = subz * Z;
    A -= subz;
    C -= subz;
    return dfs(A, B, C, X, Y, Z, 7) + 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, 0LL);
    B -= suby, C -= suby;
    cost += suby * Y;
    lint subx = max(B - C, 0LL);
    A -= subx, B -= subx;
    cost += subx * X;
    return dfs(A, B, C, X, Y, Z, 7) + 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;
    }
}
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