Submission #1131464

#	Time	Username	Problem	Language	Result	Execution time	Memory
1131464		RedGrey1993	Zoltan (COCI16_zoltan)	C++20	140 / 140	262 ms	13352 KiB

zoltan

#include <bits/stdc++.h>
 
using namespace std;
 
template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &pa) { is >> pa.first >> pa.second; return is; }
template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << "(" << pa.first << "," << pa.second << ")"; return os; }
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 multiset<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 unordered_multiset<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; 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; }
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 T> void resize_array(vector<T> &vec, int len) { vec.resize(len); }
template <typename T, typename... Args> void resize_array(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) resize_array(v, args...); }
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); }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
mt19937 mrand(random_device{}());
int rnd(int x) { return mrand() % x; }

// 函数名后面加 ll 就可以计算 long long类型对应的结果
// __builtin_ffs(x)  // 返回x中最后一个为1的位是从后向前的第几位
// __builtin_popcount(x)  // x中1的个数
// __builtin_ctz(x)  // x末尾0的个数。x=0时结果未定义。
// __builtin_clz(x)  // x前导0的个数。x=0时结果未定义。
// __builtin_parity(x)  // x中1的奇偶性。
#define highest_bit1_index(x) (31 - __builtin_clz(x))
#define highest_bit1_index_ll(x) (63 - __builtin_clzll(x))
#define rep(i, a, n) for (int i = a; i < (n); i++)
#define per(i, a, n) for (int i = (n)-1; i >= a; i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
typedef vector<int> vi;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef double db;
#if DEBUG
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
#else
#define dbg(x)
#endif

// For example: SegmentTree<int> st(n, [](int a, int b) { return max(a, b); });
template <typename T>
class SegmentTree {
 public:
  SegmentTree(int node_num, function<T(T, T)> f, T default_v = T())
      : f_(f), default_value_(default_v) {
    node_num_ = node_num;
    offset_ = 1;
    while (offset_ < node_num) offset_ *= 2;
    nodes_.resize(offset_ * 2, default_v);
  }

  friend istream& operator>>(istream& in, SegmentTree<T>& st) {
    for (int i = 0; i < st.node_num_; ++i) {
        cin >> st.nodes_[i + st.offset_ - 1];
    }
    return in;
  }

  // Init 前需要先赋初始值给nodes_
  // rep(i,0,n) {
  //     nodes_[i + offset_ - 1] = A[i];
  // }
  void Init() {
    for (int k = offset_ - 2; k >= 0; --k) {
        nodes_[k] = f_(nodes_[k * 2 + 1], nodes_[k * 2 + 2]);
    }
  }
 
  // 将第k个值更新为a
  void Update(int k, T a) {
    // 叶子节点
    k += offset_ - 1;
    // nodes_[k] = a;
    if (nodes_[k].first == a.first) nodes_[k].second += a.second;
    else nodes_[k] = a;
    // 向上更新
    while (k > 0) {
      k = (k - 1) / 2;
      nodes_[k] = f_(nodes_[k * 2 + 1], nodes_[k * 2 + 2]);
    }
  }
 
  // 迭代写法，效率更高（多找几道题验证下）
  // https://oj.uz/problem/view/balkan11_trapezoid 验证通过 迭代耗时：366 ms，递归耗时：379 ms
  // https://www.spoj.com/problems/QTREE/ 验证通过 迭代耗时：400 ms，递归耗时：700 ms
//   T Query(int lo, int hi) {
//     T ra = default_value_, rb = default_value_;
//     for (lo += offset_ - 1, hi += offset_ - 1; lo < hi; lo /= 2, hi /= 2) {
//       if (!(lo & 1)) ra = f_(ra, nodes_[lo++]);
//       if (!(hi & 1)) rb = f_(rb, nodes_[--hi]);
//     }
//     return f_(ra, rb);
//   }
  // 递归写法，更好理解
  T Query(int a, int b) { return Query(a, b, 0, 0, offset_); }
 
 private:
  // 求[a,b)区间的目标值
  // 后面的参数是为了计算方便传入的
  // k是节点的编号，l,r表示这个节点代表区间[l,r)
  // 外部调用时，使用query(a,b,0,0,n);
  T Query(int a, int b, int k, int l, int r) {
    // 如果[a,b)和[l,r)不相交，返回0
    if (r <= a || b <= l) return default_value_;
    // 如果[a,b)完全包含[l,r)，返回当前节点值
    if (a <= l && r <= b)
      return nodes_[k];
    else {
      T v1 = Query(a, b, k * 2 + 1, l, (l + r) / 2);
      T v2 = Query(a, b, k * 2 + 2, (l + r) / 2, r);
      return f_(v1, v2);
    }
  }
  int node_num_;
  int offset_;
  vector<T> nodes_;
  function<T(T, T)> f_;
  T default_value_;
};

// Right: -Mint(1), Wrong: Mint(-1) !!!
template <unsigned int MOD>
class ModInt {
public:
    ModInt(unsigned long long _v = 0) { set_v((_v % MOD + MOD)); }
    explicit operator bool() const { return val != 0; }
    ModInt operator-() const { return ModInt() - *this; }
    ModInt operator+(const ModInt &r) const { return ModInt().set_v(val + r.val); }
    ModInt operator-(const ModInt &r) const { return ModInt().set_v(val + MOD - r.val); }
    ModInt operator*(const ModInt &r) const { return ModInt().set_v((unsigned int)((unsigned long long)(val)*r.val % MOD)); }
    ModInt operator/(const ModInt &r) const { return *this * r.inv(); }
    ModInt &operator+=(const ModInt &r) { return *this = *this + r; }
    ModInt &operator-=(const ModInt &r) { return *this = *this - r; }
    ModInt &operator*=(const ModInt &r) { return *this = *this * r; }
    ModInt &operator/=(const ModInt &r) { return *this = *this / r; }
    // ModInt &operator=(unsigned long long _v) { set_v((_v % MOD + MOD)); return *this; }
    unsigned int operator=(unsigned long long _v) { set_v((_v % MOD + MOD)); return val; }
    bool operator==(const ModInt &r) const { return val == r.val; }
    bool operator!=(const ModInt &r) const { return val != r.val; }
    ModInt pow(long long n) const {
        ModInt x = *this, r = 1;
        while (n) {
            if (n & 1)
                r *= x;
            x *= x;
            n >>= 1;
        }
        return r;
    }
    ModInt inv() const { return pow(MOD - 2); }
    friend ostream &operator<<(ostream &os, const ModInt &r) { return os << r.val; }
    friend istream &operator>>(istream &is, ModInt &r) { return is >> r.val; }
private:
    unsigned int val;
    ModInt &set_v(unsigned int _v) {
        val = (_v < MOD) ? _v : _v - MOD;
        return *this;
    }
};
using Mint = ModInt<(unsigned int)(1e9+7)>;

class Solution {
public:
    void Solve() {
        int n;
        while(cin>>n) {
            vector<int> a(n);
            cin>>a;
            
            // 路径压缩
            vector<int> b = a;
            sort(all(b));
            b.erase(unique(all(b)), b.end());
            for(int i = 0; i < n; i++) {
                a[i] = lower_bound(all(b), a[i]) - b.begin();
                assert(a[i] >= 0 && a[i] < n);
            }

            vector<pair<int,Mint>> ans1(n),ans2(n);
            SegmentTree<pair<int,Mint>> st1(n, [](const pair<int,Mint>& a, const pair<int,Mint>& b) {
                if (a.first > b.first) return a;
                else if (a.first < b.first) return b;
                else return mp(a.first, a.second + b.second);
            }, mp(0,0));
            per(i,0,n) {
                auto max_len = st1.Query(a[i]+1, n);
                if (max_len.first == 0) max_len.second = 1;
                max_len.first++;
                ans1[i] = max_len;
                st1.Update(a[i], max_len);
            }

            SegmentTree<pair<int,Mint>> st2(n, [](const pair<int,Mint>& a, const pair<int,Mint>& b) {
                if (a.first > b.first) return a;
                else if (a.first < b.first) return b;
                else return mp(a.first, a.second + b.second);
            }, mp(0,0));
            per(i,0,n) {
                auto max_len = st2.Query(0, a[i]);
                if (max_len.first == 0) max_len.second = 1;
                max_len.first++;
                ans2[i] = max_len;
                st2.Update(a[i], max_len);
            }

            int max_l = 0;
            Mint max_cnt = 0;
            rep(i,0,n) {
                if (ans1[i].first + ans2[i].first - 1 > max_l) {
                    max_l = ans1[i].first + ans2[i].first - 1;
                    max_cnt = ans1[i].second * ans2[i].second * Mint(2).pow(n - max_l);
                } else if (ans1[i].first + ans2[i].first - 1 == max_l) {
                    max_cnt = (max_cnt + ans1[i].second * ans2[i].second * Mint(2).pow(n - max_l));
                }
            }

            cout << max_l << " " << max_cnt << "\n";
        }
        cout.flush();
    }
private:
};

// # define FILE_IO 1
void set_io(const string &name = "") {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
#if FILE_IO
    if (!name.empty()) {
        freopen((name + ".in").c_str(), "r", stdin);
        freopen((name + ".out").c_str(), "w", stdout);
    }
#endif
}

int main() {
    set_io("tmp");
    Solution().Solve();

    return 0;
}

• Subtask 1: 140 / 140