Submission #537707

#	Time	Username	Problem	Language	Result	Execution time	Memory
537707		cig32	Hedgehog Daniyar and Algorithms (IZhO19_sortbooks)	C++17	0 / 100	3054 ms	40104 KiB

sortbooks

#include "bits/stdc++.h"
using namespace std;
const int MAXN = 2e5 + 10;
const int MOD = 1e9 + 7;
//#define int long long
 
mt19937_64 rng((int)std::chrono::steady_clock::now().time_since_epoch().count());
int rnd(int x, int y) {
  int u = uniform_int_distribution<int>(x, y)(rng); return u;
}
long long bm(long long b, long long p) { 
  if(p==0) return 1;
  long long r = bm(b, p/2);
  if(p&1) return (((r*r) % MOD) * b) % MOD;
  return (r*r) % MOD;
}

struct wavelet_tree{
	#define vi vector<int>
	#define pb push_back
	int lo, hi;
	wavelet_tree *l, *r;
	vi b;
 
	//nos are in range [x,y]
	//array indices are [from, to)
	wavelet_tree(int *from, int *to, int x, int y){
		lo = x, hi = y;
		if(lo == hi or from >= to) return;
		int mid = (lo+hi)/2;
		auto f = [mid](int x){
			return x <= mid;
		};
		b.reserve(to-from+1);
		b.pb(0);
		for(auto it = from; it != to; it++)
			b.pb(b.back() + f(*it));
		//see how lambda function is used here	
		auto pivot = stable_partition(from, to, f);
		l = new wavelet_tree(from, pivot, lo, mid);
		r = new wavelet_tree(pivot, to, mid+1, hi);
	}
 
	//kth smallest element in [l, r]
	int kth(int l, int r, int k){
		if(l > r) return 0;
		if(lo == hi) return lo;
		int inLeft = b[r] - b[l-1];
		int lb = b[l-1]; //amt of nos in first (l-1) nos that go in left 
		int rb = b[r]; //amt of nos in first (r) nos that go in left
		if(k <= inLeft) return this->l->kth(lb+1, rb , k);
		return this->r->kth(l-lb, r-rb, k-inLeft);
	}
 
	//count of nos in [l, r] Less than or equal to k
	int LTE(int l, int r, int k) {
		if(l > r or k < lo) return 0;
		if(hi <= k) return r - l + 1;
		int lb = b[l-1], rb = b[r];
		return this->l->LTE(lb+1, rb, k) + this->r->LTE(l-lb, r-rb, k);
	}
 
	//count of nos in [l, r] equal to k
	int count(int l, int r, int k) {
		if(l > r or k < lo or k > hi) return 0;
		if(lo == hi) return r - l + 1;
		int lb = b[l-1], rb = b[r], mid = (lo+hi)/2;
		if(k <= mid) return this->l->count(lb+1, rb, k);
		return this->r->count(l-lb, r-rb, k);
	}

  //count max element in [l, r] stricly smaller than k
  int upper_bound(int l, int r, int k) {
    if(l > r) return -1;
    int uwu = LTE(l, r, k - 1);
    if(uwu == 0) return -1;
    return kth(l, r, uwu);
  }
	~wavelet_tree(){
		delete l;
		delete r;
	}
};

int st[20][MAXN];

int query_max(int l, int r) {
  int k = 32 - __builtin_clz(r-l+1) - 1;
  return max(st[k][l], st[k][r-(1<<k)+1]);
}

int ans[20][MAXN];
int a[MAXN];
int upper_bound(int l, int r, int v) {
  if(l > r) return -1;
  int ans = 0;
  for(int i=l; i<=r; i++) {
    if(a[i] < v) ans = max(ans, a[i]);
  }
  return ans;
}

void solve(int tc) {
  int n, m;
  cin >> n >> m;
  for(int i=1; i<=n; i++) {
    cin >> a[i];
    st[0][i] = a[i];
  }
//  wavelet_tree T(a+1, a+n+1, 0, 1000000000);
  for(int i=1; i<20; i++) {
    for(int j=1; j<=n; j++) {
      if(j+(1<<i)-1 <= n) {
        st[i][j] = max(st[i-1][j], st[i-1][j+(1<<(i-1))]);
      }
    }
  }
  for(int i=1; i<=n; i++) {
    ans[0][i] = -1;
  }
  for(int i=1; i<20; i++) {
    for(int j=1; j<=n; j++) {
      if(j+(1<<i)-1 <= n) {
        ans[i][j] = max(ans[i-1][j], ans[i-1][j+(1<<(i-1))]);
        int q1 = query_max(j, j+(1<<(i-1))-1);
        int q2 = upper_bound(j + (1<<(i-1)), j + (1<<i) - 1, q1);
        if(q1 > q2 && q2 != -1) ans[i][j] = max(ans[i][j], q1 + q2);
      }
    }
  }
  while(m--) {
    int l, r, s;
    cin >> l >> r >> s;
    if(l == r) {
      cout << "1\n"; continue;
    }
    int k = 32 - __builtin_clz(r-l+1) - 1;
    int res = max(ans[k][l], ans[k][r-(1<<k)+1]);
    int q1 = query_max(l, l + (1<<k) - 1);
    int q2 = upper_bound(l + (1<<k), r, q1);
    if(q1 > q2 && q2 != -1) res = max(res, q1 + q2);
    cout << (res > s ? "0\n" : "1\n");
  }
}

int32_t main(){
  ios::sync_with_stdio(0); cin.tie(0);
  int t = 1; //cin >> t;
  for(int i=1; i<=t; i++) solve(i);
}

/*
7 1
4 9 1 0 4 0 8 
4 7 4
*/

• Subtask 1: 0 / 8
• Subtask 2: 0 / 9
• Subtask 3: 0 / 13
• Subtask 4: 0 / 17
• Subtask 5: 0 / 30
• Subtask 6: 0 / 23