제출 #197599

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
197599		stefdasca	Watering can (POI13_kon)	C++14	100 / 100	3132 ms	100828 KiB

kon

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

const int MAXN = 300001, rad = 400;

int N, K, v[MAXN+5], done = 200000;

int buk[MAXN+5], L[1012], R[1012], ad[1012], nrbuk, aans[1012];

short ans[1002][100012];

void proc(int bk, int LL, int RR)
{
    for(int j = L[bk]; j <= R[bk]; ++j)
    {
        if(v[j] >= K)
            ans[bk][0] = 0;
        else
            if(K - v[j] <= 100000)
                ans[bk][K - v[j]] = 0;
    }
    for(int j = LL; j <= RR; ++j)
        v[j]++;
    for(int j = L[bk]; j <= R[bk]; ++j)
    {
        v[j] += ad[bk];
        if(v[j] >= K)
            ans[bk][0]++;
        else
            if(K - v[j] <= 100000)
                ans[bk][K - v[j]]++;
    }
    ad[bk] = 0;
    aans[bk] = ans[bk][0];
}

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, 1, 0);
}

void podlej(int a, int b)
{
    int fini = R[buk[a]];
    int ax = buk[a];
    proc(ax, a, min(b, fini));
    a = min(b, fini) + 1;
    while(a < N && R[buk[a]] <= b)
        ++ad[buk[a]], aans[buk[a]] = aans[buk[a]] + ans[buk[a]][ad[buk[a]]], a = R[buk[a]] + 1;
    if(a <= b)
    {
        fini = R[buk[a]];
        ax = buk[a];
        proc(ax, a, b);
    }
    --done;
}

int dojrzale(int a, int b)
{
    int fini = R[buk[a]];
    int qans = 0;
    proc(buk[a], 1, 0);
    while(a <= b && a <= fini)
    {
        if(v[a] >= K)
            ++qans;
        ++a;
    }
    while(a < N && R[buk[a]] <= b)
        qans += aans[buk[a]], a = R[buk[a]] + 1;
    if(a <= b)
    {
        proc(buk[a], 1, 0);
        fini = R[buk[a]];
        while(a <= b && a <= fini)
        {
            if(v[a] >= K)
                ++qans;
            ++a;
        }
    }
    --done;
    return qans;
}

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