\nusing namespace std;\n\n#define int long long\n\nbool check3(vector
a) {\n //for (int i: a) cout << i << ' '; cout << endl;\n\tsort(a.begin(), a.end());\n\treturn a[2] < a[1] + a[0];\n}\n\nbool check6(vector a) {\n //for (int i: a) cout << i << ' '; cout << endl;\n sort(a.begin(), a.end());\n\tfor (int i = 2; i < 5; i++) {\n\t\tvector p;\n\t\tfor (int j = 0; j < 5; j++) {\n\t\t if (i != j) p.push_back(j);\n\t\t}\n\t\t\n\t\tdo {\n\t\t vector s(2); \n\t\t if (p[0] < i && p[1] < i) s[0] = a[p[0]] + a[p[1]] - a[i];\n\t\t if (p[2] < 5 && p[3] < 5) s[1] = a[p[2]] + a[p[3]] - a[5];\n\t\t \n\t\t if (s[0] > 0 && s[1] > 0) return true;\n\t\t} while (next_permutation(p.begin(), p.end()));\n\t}\n\treturn false;\n}\n\nsigned main() {\n\tios::sync_with_stdio(0); cin.tie(0);\n\n\tint n; cin >> n;\n\tvector a(n);\n\tfor (int i = 0; i < n; i++) {\n\t\tcin >> a[i];\n\t}\n\tsort(a.begin(), a.end());\n\t\n\tvector ans2;\n\tfor (int i = 0; i+5 < n; i++) {\n\t vector tmp;\n\t for (int j = 0; j < 6; j++) {\n\t tmp.push_back(a[i+j]);\n\t }\n\t if (check6(tmp)) {\n\t ans2 = tmp;\n\t }\n\t}\n\t\n\tvector ans;\n\n\twhile (a.size() >= 3) {\n\t\tans.push_back(a.back()); a.pop_back();\n\t\tans.push_back(a.back()); a.pop_back();\n\t\tans.push_back(a.back()); a.pop_back();\n\n\t\tif (check3(ans)) break;\n\t\telse {\n\t\t\ta.push_back(ans.back()); ans.pop_back();\n\t\t\ta.push_back(ans.back()); ans.pop_back();\n\t\t\tans.pop_back();\n\t\t}\n\t}\n\n\twhile (a.size() >= 3) {\n\t\tans.push_back(a.back()); a.pop_back();\n\t\tans.push_back(a.back()); a.pop_back();\n\t\tans.push_back(a.back()); a.pop_back();\n\n\t\tif (check6(ans)) break;\n\t\telse {\n\t\t\ta.push_back(ans.back()); ans.pop_back();\n\t\t\ta.push_back(ans.back()); ans.pop_back();\n\t\t\tans.pop_back();\n\t\t}\n\t}\n\n\tcout << max(accumulate(ans.begin(), ans.end(), 0LL), accumulate(ans2.begin(), ans2.end(), 0LL)) << '\\n';\n\n\treturn 0;\n}","accuracy":1,"time_ms":20,"memory_kb":3880,"score_of_the_acc":-0.3994,"final_rank":11},{"submission_id":"aoj_2265_10042798","code_snippet":"#include \nusing namespace std;\n\ntypedef long long ll;\n\nint main(){\n ios::sync_with_stdio(false);\n cin.tie(0);\n \n ll N;\n cin >> N;\n vector a(N);\n for(auto &x: a) cin >> x;\n \n // Сортируем в неубывающем порядке\n sort(a.begin(), a.end(), [](const ll &x, const ll &y) -> bool{\n return x > y;\n });\n \n // Если палочек меньше 6, выводим 0\n if(N < 6){\n cout << 0 << endl;\n return 0;\n }\n \n // Выбираем первые K палочек\n ll K = min((ll)30, N);\n vector topK(a.begin(), a.begin() + K);\n \n // Перебираем все возможные тройки и сохраняем валидные\n struct Triple {\n int i, j, k;\n ll sum;\n };\n \n vector triples;\n \n for(int i = 0; i < K-2; ++i){\n for(int j = i+1; j < K-1; ++j){\n for(int k = j+1; k < K; ++k){\n if(topK[j] + topK[k] > topK[i]){\n triples.push_back(Triple{ i, j, k, topK[i] + topK[j] + topK[k] });\n }\n }\n }\n }\n \n // Если меньше двух тройк, невозможно\n if(triples.size() < 2){\n // Возможно, стоит проверить дальше, но для упрощения ограничимся K=30\n cout << 0 << endl;\n return 0;\n }\n \n // Ищем максимальную сумму двух не пересекающихся тройк\n ll max_sum = 0;\n for(int i = 0; i < triples.size(); ++i){\n for(int j = i+1; j < triples.size(); ++j){\n // Проверяем, не пересекаются ли тройки\n int a1 = triples[i].i;\n int b1 = triples[i].j;\n int c1 = triples[i].k;\n \n int a2 = triples[j].i;\n int b2 = triples[j].j;\n int c2 = triples[j].k;\n \n // Проверяем уникальность индексов\n if(a1 != a2 && a1 != b2 && a1 != c2 &&\n b1 != a2 && b1 != b2 && b1 != c2 &&\n c1 != a2 && c1 != b2 && c1 != c2){\n ll total = triples[i].sum + triples[j].sum;\n if(total > max_sum){\n max_sum = total;\n }\n }\n }\n }\n \n // Выводим результат\n cout << max_sum << endl;\n}","accuracy":0.136986301369863,"time_ms":10,"memory_kb":3592,"score_of_the_acc":-0.3261,"final_rank":19},{"submission_id":"aoj_2265_10042791","code_snippet":"#include \nusing namespace std;\n\ntypedef long long ll;\n\nint main(){\n ios::sync_with_stdio(false);\n cin.tie(0);\n \n ll N;\n cin >> N;\n vector a(N);\n for(auto &x: a) cin >> x;\n \n // Сортируем в неубывающем порядке\n sort(a.begin(), a.end(), [](const ll &x, const ll &y) -> bool{\n return x > y;\n });\n \n // Если палочек меньше 6, выводим 0\n if(N < 6){\n cout << \"0\";\n return 0;\n }\n \n // Выбираем первые K палочек\n ll K = min((ll)30, N);\n vector topK(a.begin(), a.begin() + K);\n \n // Перебираем все возможные тройки и сохраняем валидные\n struct Triple {\n int i, j, k;\n ll sum;\n };\n \n vector triples;\n \n for(int i = 0; i < K-2; ++i){\n for(int j = i+1; j < K-1; ++j){\n for(int k = j+1; k < K; ++k){\n if(topK[j] + topK[k] > topK[i]){\n triples.push_back(Triple{ i, j, k, topK[i] + topK[j] + topK[k] });\n }\n }\n }\n }\n \n // Если меньше двух тройк, невозможно\n if(triples.size() < 2){\n // Возможно, стоит проверить дальше, но для упрощения ограничимся K=30\n cout << \"0\";\n return 0;\n }\n \n // Ищем максимальную сумму двух не пересекающихся тройк\n ll max_sum = 0;\n for(int i = 0; i < triples.size(); ++i){\n for(int j = i+1; j < triples.size(); ++j){\n // Проверяем, не пересекаются ли тройки\n int a1 = triples[i].i;\n int b1 = triples[i].j;\n int c1 = triples[i].k;\n \n int a2 = triples[j].i;\n int b2 = triples[j].j;\n int c2 = triples[j].k;\n \n // Проверяем уникальность индексов\n if(a1 != a2 && a1 != b2 && a1 != c2 &&\n b1 != a2 && b1 != b2 && b1 != c2 &&\n c1 != a2 && c1 != b2 && c1 != c2){\n ll total = triples[i].sum + triples[j].sum;\n if(total > max_sum){\n max_sum = total;\n }\n }\n }\n }\n \n // Выводим результат\n cout << max_sum << endl;\n}","accuracy":0.1232876712328767,"time_ms":10,"memory_kb":3588,"score_of_the_acc":-0.3254,"final_rank":20},{"submission_id":"aoj_2265_9128309","code_snippet":"#pragma GCC optimize(\"Ofast\")\n#include \nusing namespace std;\ntypedef long long int ll;\ntypedef unsigned long long int ull;\n\nmt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());\nll myRand(ll B) { return (ull)rng() % B; }\n\nint main() {\n cin.tie(nullptr);\n ios::sync_with_stdio(false);\n int n;\n cin >> n;\n vector a(n);\n for (int i = 0; i < n; i++) {\n cin >> a[i];\n }\n sort(a.begin(), a.end());\n ll res = 0;\n auto can = [&](ll x, ll y, ll z) -> ll {\n if (x + y > z) return x + y + z;\n else return -1e18;\n };\n {\n ll mx = -1e18;\n for (int i = 5; i < n; i++) {\n mx = max(mx, can(a[i - 5], a[i - 4], a[i - 3]));\n ll u = can(a[i - 2], a[i - 1], a[i]);\n res = max(res, u + mx);\n }\n }\n {\n for (int i = 0; i + 5 < n; i++) {\n res = max(res, can(a[i], a[i + 1], a[i + 3]) + can(a[i + 2], a[i + 4], a[i + 5]));\n res = max(res, can(a[i], a[i + 1], a[i + 4]) + can(a[i + 2], a[i + 3], a[i + 5]));\n res = max(res, can(a[i], a[i + 1], a[i + 5]) + can(a[i + 2], a[i + 3], a[i + 4]));\n res = max(res, can(a[i], a[i + 2], a[i + 3]) + can(a[i + 1], a[i + 4], a[i + 5]));\n res = max(res, can(a[i], a[i + 2], a[i + 4]) + can(a[i + 1], a[i + 3], a[i + 5]));\n res = max(res, can(a[i], a[i + 2], a[i + 5]) + can(a[i + 1], a[i + 3], a[i + 4]));\n res = max(res, can(a[i], a[i + 3], a[i + 4]) + can(a[i + 1], a[i + 2], a[i + 5]));\n res = max(res, can(a[i], a[i + 3], a[i + 5]) + can(a[i + 1], a[i + 2], a[i + 4]));\n res = max(res, can(a[i], a[i + 4], a[i + 5]) + can(a[i + 1], a[i + 2], a[i + 3]));\n }\n }\n cout << res << endl;\n}","accuracy":1,"time_ms":10,"memory_kb":3896,"score_of_the_acc":-0.3811,"final_rank":9},{"submission_id":"aoj_2265_8965525","code_snippet":"#include \nusing namespace std;\nusing ll=int64_t;\n\nint main() {\n int n;\n cin >> n;\n if(n<6){\n cout << 0 << endl;\n return 0;\n }\n \n priority_queue pq;\n ll a;\n \n for(int i=0;i> a;\n pq.push(a);\n }\n \n ll ans1=0,ans=0;\n deque san; \n int cnt=0,num=0;\n while(!pq.empty()){\n //cout << pq.size() << \" \" << cnt << \" \" << num << endl;\n if(num==0){\n if(cnt<3){ \n san.push_back(pq.top());\n pq.pop();\n cnt++;\n }\n if(cnt==3){\n if(san[0]0) break;\n }\n \n cout << max(ans,ans1) << endl;\n}","accuracy":1,"time_ms":30,"memory_kb":4204,"score_of_the_acc":-0.4793,"final_rank":12},{"submission_id":"aoj_2265_5114777","code_snippet":"#include \n//#include \n//using namespace atcoder;\n#pragma GCC target (\"avx2\")\n#pragma GCC optimization (\"O3\")\n#pragma GCC optimization (\"unroll-loops\")\n\nusing namespace std;\n\ntypedef vector VI;\ntypedef vector VVI;\ntypedef vector VS;\ntypedef pair PII;\ntypedef pair pii;\ntypedef pair PLL;\ntypedef pair TIII;\n\ntypedef long long ll;\ntypedef long double ld;\ntypedef unsigned long long ull;\n\n\n#define FOR(i, s, n) for (int i = s; i < (int)n; ++i)\n#define REP(i, n) FOR(i, 0, n)\n#define rep(i, a, b) for (int i = a; i < (b); ++i)\n#define trav(a, x) for (auto &a : x)\n#define all(x) x.begin(), x.end()\n\n#define MOD 1000000007\n\ntemplate inline bool chmax(T1 &a, T2 b) {if (a < b) {a = b; return true;} return false;}\ntemplate inline bool chmin(T1 &a, T2 b) {if (a > b) {a = b; return true;} return false;}\nconst double EPS = 1e-9, PI = acos(-1);\nconst double pi = 3.141592653589793238462643383279;\n \ntypedef string::const_iterator State;\nll GCD(ll a, ll b){\n return (b==0)?a:GCD(b, a%b);\n}\nll LCM(ll a, ll b){\n return a/GCD(a, b) * b;\n}\ntemplate< int mod >\nstruct ModInt {\n int x;\n\n ModInt() : x(0) {}\n\n ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}\n\n ModInt &operator+=(const ModInt &p) {\n if((x += p.x) >= mod) x -= mod;\n return *this;\n }\n\n ModInt &operator-=(const ModInt &p) {\n if((x += mod - p.x) >= mod) x -= mod;\n return *this;\n }\n\n ModInt &operator*=(const ModInt &p) {\n x = (int) (1LL * x * p.x % mod);\n return *this;\n }\n\n ModInt &operator/=(const ModInt &p) {\n *this *= p.inverse();\n return *this;\n }\n\n ModInt operator-() const { return ModInt(-x); }\n\n ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }\n\n ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }\n\n ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }\n\n ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }\n\n bool operator==(const ModInt &p) const { return x == p.x; }\n\n bool operator!=(const ModInt &p) const { return x != p.x; }\n\n ModInt inverse() const {\n int a = x, b = mod, u = 1, v = 0, t;\n while(b > 0) {\n t = a / b;\n swap(a -= t * b, b);\n swap(u -= t * v, v);\n }\n return ModInt(u);\n }\n\n ModInt pow(int64_t n) const {\n ModInt ret(1), mul(x);\n while(n > 0) {\n if(n & 1) ret *= mul;\n mul *= mul;\n n >>= 1;\n }\n return ret;\n }\n\n friend ostream &operator<<(ostream &os, const ModInt &p) {\n return os << p.x;\n }\n\n friend istream &operator>>(istream &is, ModInt &a) {\n int64_t t;\n is >> t;\n a = ModInt< mod >(t);\n return (is);\n }\n\n static int get_mod() { return mod; }\n};\n\nusing modint = ModInt< 998244353 >;\ntemplate< typename T >\nstruct Combination {\n vector< T > _fact, _rfact, _inv;\n\n Combination(int sz) : _fact(sz + 1), _rfact(sz + 1), _inv(sz + 1) {\n _fact[0] = _rfact[sz] = _inv[0] = 1;\n for(int i = 1; i <= sz; i++) _fact[i] = _fact[i - 1] * i;\n _rfact[sz] /= _fact[sz];\n for(int i = sz - 1; i >= 0; i--) _rfact[i] = _rfact[i + 1] * (i + 1);\n for(int i = 1; i <= sz; i++) _inv[i] = _rfact[i] * _fact[i - 1];\n }\n\n inline T fact(int k) const { return _fact[k]; }\n\n inline T rfact(int k) const { return _rfact[k]; }\n\n inline T inv(int k) const { return _inv[k]; }\n\n T P(int n, int r) const {\n if(r < 0 || n < r) return 0;\n return fact(n) * rfact(n - r);\n }\n\n T C(int p, int q) const {\n if(q < 0 || p < q) return 0;\n return fact(p) * rfact(q) * rfact(p - q);\n }\n\n T H(int n, int r) const {\n if(n < 0 || r < 0) return (0);\n return r == 0 ? 1 : C(n + r - 1, r);\n }\n};\nstruct UnionFind{\n vector par;\n vector siz;\n\n UnionFind(int sz_): par(sz_), siz(sz_) {\n for(int i=0; i> N;\n vector a(N);\n REP(i,N) cin >> a[i];\n sort(all(a));\n ll t1=-1, t2=-1;\n ll ans = 0;\n for(int i=N-1; i>=2; i--){\n if(used[i]) continue;\n ll e1 = a[i], e2 = a[i-1], e3 = a[i-2];\n if(e2+e3 > e1) {\n if(t1 == -1){\n t1 = e1+e2+e3;\n used[i] = used[i-1] = used[i-2] = true;\n }else{\n t2 = e1+e2+e3;\n break;\n }\n }\n }\n if(t2 != -1) ans = t1+t2;\n for(int i=N-1; i>=5; i--){\n vector v{a[i-5], a[i-4], a[i-3], a[i-2], a[i-1], a[i]};\n ll sum = accumulate(all(v), 0LL);\n\n vector ok(1<<6);\n REP(bit,(1<<6)){\n if(__builtin_popcount(bit) != 3) continue;\n vector e;\n for(int j=0; j<6; j++){\n if(bit >> j & 1) e.push_back(v[j]);\n }\n\n if(e[0]+e[1]>e[2] && e[0]+e[2]>e[1] && e[1]+e[2]>e[0]){\n ok[bit] = true;\n }\n }\n REP(bit,(1<<6)){\n if(__builtin_popcount(bit) != 3) continue;\n if(ok[bit] && ok[((1<<6)-1) & ~bit]){\n ans = max(ans, sum);\n break;\n } \n }\n }\n cout << ans << endl;\n return 0;\n}","accuracy":1,"time_ms":180,"memory_kb":3624,"score_of_the_acc":-0.6936,"final_rank":15},{"submission_id":"aoj_2265_4796231","code_snippet":"#include \nusing namespace std;\n\n#define int long long\n\nbool check3(vector a) {\n //for (int i: a) cout << i << ' '; cout << endl;\n\tsort(a.begin(), a.end());\n\treturn a[2] < a[1] + a[0];\n}\n\nbool check6(vector a) {\n //for (int i: a) cout << i << ' '; cout << endl;\n sort(a.begin(), a.end());\n\tfor (int i = 2; i < 5; i++) {\n\t\tvector p;\n\t\tfor (int j = 0; j < 5; j++) {\n\t\t if (i != j) p.push_back(j);\n\t\t}\n\t\t\n\t\tdo {\n\t\t vector s(2); \n\t\t s[0] = a[p[0]] + a[p[1]] - a[i];\n\t\t s[1] = a[p[2]] + a[p[3]] - a[5];\n\t\t \n\t\t if (s[0] > 0 && s[1] > 0) return true;\n\t\t} while (next_permutation(p.begin(), p.end()));\n\t}\n\treturn false;\n}\n\nsigned main() {\n\tios::sync_with_stdio(0); cin.tie(0);\n\n\tint n; cin >> n;\n\tvector a(n);\n\tfor (int i = 0; i < n; i++) {\n\t\tcin >> a[i];\n\t}\n\tsort(a.begin(), a.end());\n\t\n\tvector ans2;\n\tfor (int i = 0; i+5 < n; i++) {\n\t vector tmp;\n\t for (int j = 0; j < 6; j++) {\n\t tmp.push_back(a[i+j]);\n\t }\n\t if (check6(tmp)) {\n\t ans2 = tmp;\n\t }\n\t}\n\t\n\tvector ans;\n\n\twhile (a.size() >= 3) {\n\t\tans.push_back(a.back()); a.pop_back();\n\t\tans.push_back(a.back()); a.pop_back();\n\t\tans.push_back(a.back()); a.pop_back();\n\n\t\tif (check3(ans)) break;\n\t\telse {\n\t\t\ta.push_back(ans.back()); ans.pop_back();\n\t\t\ta.push_back(ans.back()); ans.pop_back();\n\t\t\tans.pop_back();\n\t\t}\n\t}\n\n\twhile (a.size() >= 3) {\n\t\tans.push_back(a.back()); a.pop_back();\n\t\tans.push_back(a.back()); a.pop_back();\n\t\tans.push_back(a.back()); a.pop_back();\n\n\t\tif (check6(ans)) break;\n\t\telse {\n\t\t\ta.push_back(ans.back()); ans.pop_back();\n\t\t\ta.push_back(ans.back()); ans.pop_back();\n\t\t\tans.pop_back();\n\t\t}\n\t}\n\n\tcout << max(accumulate(ans.begin(), ans.end(), 0LL), accumulate(ans2.begin(), ans2.end(), 0LL)) << '\\n';\n\n\treturn 0;\n}","accuracy":0.6438356164383562,"time_ms":30,"memory_kb":3552,"score_of_the_acc":-0.3614,"final_rank":18},{"submission_id":"aoj_2265_2660468","code_snippet":"#include \n#include \n#include \n#include \n#include \n#include \n#include \n#include \n#include \n#include \n#include