結果

問題 No.966 引き算をして門松列（その１）
ユーザー fumiphysfumiphys
提出日時 2020-01-13 20:34:15
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 23 ms / 2,000 ms
コード長 3,718 bytes
コンパイル時間 1,623 ms
コンパイル使用メモリ 172,628 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-01 10:39:22
合計ジャッジ時間 2,184 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
（要ログイン）

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
6,812 KB
testcase_01 AC 2 ms
6,944 KB
testcase_02 AC 2 ms
6,940 KB
testcase_03 AC 2 ms
6,940 KB
testcase_04 AC 17 ms
6,940 KB
testcase_05 AC 18 ms
6,940 KB
testcase_06 AC 23 ms
6,940 KB
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 # 

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define whole(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;

bool ifk(vector<ll> v){
  if(v[0] == v[1] || v[1] == v[2] || v[2] == v[0])return false;
  if(v[0] > v[1] && v[1] > v[2])return false;
  if(v[0] < v[1] && v[1] < v[2])return false;
  return true;
}

int main(int argc, char const* argv[])
{
  int t; cin >> t;
  rep(i_, t){
    vector<ll> v(3); cin >> v;
    vector<int> e(3); rep(i, 3)e[i] = i;
    ll res = linf;
    do{
      ll tmp = 0;
      vector<ll> V = v;
      if(V[e[0]] <= 2 || V[e[1]] <= 1)continue;
      if(V[e[0]] <= V[e[1]]){
        tmp += V[e[1]] - (V[e[0]] - 1);
        V[e[1]] = V[e[0]] - 1;
      }
      if(V[e[1]] <= V[e[2]]){
        tmp += V[e[2]] - (V[e[1]] - 1);
        V[e[2]] = V[e[1]] - 1;
      }
      if(ifk(V))res = min(res, tmp);
    }while(next_permutation(whole(e)));
    if(res == linf)cout << -1 << endl;
    else cout << res << endl;
  }
  return 0;
}
0