Submission #197580

#	Time	Username	Problem	Language	Result	Execution time	Memory
197580		stefdasca	Watering can (POI13_kon)	C++14	40 / 100	4101 ms	11316 KiB

kon

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

const int MAXN = 300001, rad = 550;

int N, K, v[MAXN+5];

int buk[MAXN+5], L[1002], R[1002], ad[1002], nrbuk;

int srt[1002][1002];


void proc(int bk)
{
    for(int j = L[bk]; j <= R[bk]; ++j)
    {
        v[j] += ad[bk];
        srt[bk][j - L[bk]] = v[j];
    }
    ad[bk] = 0;
    sort(srt[bk], srt[bk] + R[bk] - L[bk] + 1);
}

int solve(int bk)
{
    if(srt[bk][R[bk] - L[bk]] + ad[bk] < K)
        return 0;
    int st = 0;
    int dr = R[bk] - L[bk];
    int ans = dr;
    while(st <= dr)
    {
        int mid = (st + dr) / 2;
        if(srt[bk][mid] + ad[bk] >= K)
            ans = mid, dr = mid - 1;
        else
            st = mid + 1;
    }
    return (R[bk] - L[bk]) - ans + 1;
}

void inicjuj(int n, int k, int *D)
{
    N = n, K = k;
    for(int i = 0; i < n; ++i)
        v[i] = D[i];
    for(int i = 0; i < n; ++i)
    {
        if(i % rad == 0)
            ++nrbuk, L[nrbuk] = i;
        R[nrbuk] = i;
        buk[i] = nrbuk;
    }
    for(int i = 0; i <= nrbuk; ++i)
        proc(i);
}

void podlej(int a, int b)
{
    int fini = R[buk[a]];
    while(a <= b && a <= fini)
    {
        ++v[a];
        ++a;
    }
    proc(buk[a]);
    while(a < N && R[buk[a]] <= b)
        ++ad[buk[a]], a = R[buk[a]] + 1;
    if(a <= b)
    {
        fini = R[buk[a]];
        while(a <= b && a <= fini)
        {
            ++v[a];
            ++a;
        }
        proc(buk[a]);
    }
}

int dojrzale(int a, int b)
{
    int fini = R[buk[a]];
    int ans = 0;
  //  cout << a << " " << fini << '\n';
    proc(buk[a]);
    while(a <= b && a <= fini)
    {
        if(v[a] >= K)
            ++ans;
        ++a;
    }
   // cout << "RC " << a << " " << ans << '\n';
    while(a < N && R[buk[a]] <= b)
        ans += solve(buk[a]), a = R[buk[a]] + 1;
   // cout << "RC " << a << " " << ans << '\n';
    if(a <= b)
    {
        proc(buk[a]);
        fini = R[buk[a]];
        while(a <= b && a <= fini)
        {
            if(v[a] >= K)
                ++ans;
            ++a;
        }
    }
    return ans;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10