제출 #420204

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
420204		Aldas25	말 (IOI15_horses)	C++14	77 / 100	1546 ms	40824 KiB

horses

#pragma GCC optimize ("O2")
#pragma GCC optimize ("unroll-loops")

#include "horses.h"
#include <bits/stdc++.h>

using namespace std;

#define FAST_IO ios_base::sync_with_stdio(0); cin.tie(nullptr)
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define REP(n) FOR(O, 1, (n))
#define f first
#define s second
#define pb push_back
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vii;
typedef vector<ll> vl;

const int MAXN = 500100;
const ll MOD = 1e9+7;

int n;
ll x[MAXN], y[MAXN];

ll tree3[4*MAXN];

void updPref (int p, ll val, int id = 1, int le = 1, int ri = n) {
    if (le == ri){
        tree3[id] = val;
        return;
    }
    int mid = (le+ri)/2;
    if (p <= mid)
        updPref(p, val, 2*id, le, mid);
    else
        updPref(p, val, 2*id+1, mid+1, ri);
    tree3[id] = (tree3[2*id] * tree3[2*id+1])%MOD;
}

ll prefX (int fr, int to, int id = 1, int le = 1, int ri = n) {
    if (fr > ri || to < le) return 1;
    if (fr <= le && ri <= to) return tree3[id];
    int mid = (le+ri)/2;
    return (prefX(fr, to, 2*id, le, mid) * prefX(fr, to, 2*id+1, mid+1, ri))%MOD;
}

ll prefX (int p) {
    return prefX(1, p);
}

ll tree2[4*MAXN];

void upd2 (int p, ll val, int id = 1, int le = 1, int ri = n) {
    if (le == ri){
        tree2[id] = val;
        return;
    }
    int mid = (le+ri)/2;
    if (p <= mid)
        upd2(p, val, 2*id, le, mid);
    else
        upd2(p, val, 2*id+1, mid+1, ri);
    tree2[id] = tree2[2*id] + tree2[2*id+1];
}

ll get2 (int fr, int to, int id = 1, int le = 1, int ri = n) {
    if (fr > ri || to < le) return 0;
    if (fr <= le && ri <= to) return tree2[id];
    int mid = (le+ri)/2;
    return get2(fr, to, 2*id, le, mid) + get2(fr, to, 2*id+1, mid+1, ri);
}

ll get2 (int p) {
    return get2(1, p);
}

int getId (ll sum, int id = 1, int le = 1, int ri = n) {
    if (le == ri) return le;
    int mid = (le+ri)/2;
    if (sum <= tree2[2*id]) return getId(sum, 2*id, le, mid);
    else return getId(sum-tree2[2*id], 2*id+1, mid+1, ri);
}

ll tree[4*MAXN];
void updTree (int p, ll val, int id =  1, int le = 1, int ri = n) {
    if (le == ri){
        tree[id] = le;
        return;
    }
    int mid = (le+ri)/2;
    if (p <= mid)
        updTree(p, val, 2*id, le, mid);
    else
        updTree(p, val, 2*id+1, mid+1, ri);
    if (y[tree[2*id]] > y[tree[2*id+1]])
        tree[id] = tree[2*id];
    else
        tree[id] = tree[2*id+1];
}

ll getMaxY (int fr, int to, int id = 1, int le = 1, int ri = n) {
    if (fr > ri || to < le) return 0;
    if (fr <= le && ri <= to) return tree[id];
    int mid = (le+ri)/2;
    ll id1 = getMaxY(fr, to, 2*id, le, mid);
    ll id2 = getMaxY(fr, to, 2*id+1, mid+1, ri);
    if (y[id1] > y[id2]) return id1;
    else return id2;
}

void print (){
    FOR(i, 1, n) {
        cout << " i = " << i << endl;
        cout << "    x = " << x[i] << " y = " << y[i] << endl;
        cout << "    prefX = " << prefX(i) << endl;
        cout << "    get2 pref = " << get2(i) << " get2 i = " << get2(i,i) << endl;
        cout << "    maxY i = " << getMaxY(i, i) << "  getMaxY pref = " << getMaxY(1, i) << endl;
    }
}

ll getMax () {
    int id = 1;
    ll tot = get2(1,n);
    if (tot > 35) {
        id = getId(tot-32);
    }

    ll cr = 1;
    FOR(i, id+1, n) {

        //cout << " i = " << i << " id = " << id << endl;

        if (x[i] > 1) {
            cr *= x[i];
            if (cr * y[i] > y[id]) {
                id = i;
                cr = 1;
            }
            //cout << "   case xi > 1, id = " << id << endl;
        } else {
            int le = i;
            if (get2(le,n) == 0) le = n;
            else {
                ll sm = (i > 1) ? get2(1, i-1) : 0;
                sm++;
                le = getId(sm)-1;
            }
            /// [i;le] all x are equal to 1
            int mxId = getMaxY(i, le);
        //    cout << "   case xi = 1,  le = " << le << endl;
            if (cr * y[mxId] > y[id]) {
                id = mxId;
                cr = 1;
            }
            i = le;
      //      cout << "       id = " << id << " i = " << i << endl;
        }
    }

    //cout << " final id = " << id << endl;

    ll ret = (prefX(id) * y[id])%MOD;

    //cout << "  prefX = " << prefX(id) << " y = " << y[id] << endl;
    //print();

    return ret;
}

int init(int N, int X[], int Y[]) {
    n = N;
    FOR(i, 1, n) x[i] = X[i-1];
    FOR(i, 1, n) y[i] = Y[i-1];
//    FOR(i, 0, MAXN-1) fen[i] = 1;
    FOR(i, 1, n) updPref (i, x[i]);
    FOR(i, 1, n) upd2(i, (x[i] > 1));
    FOR(i, 1, n) updTree(i, y[i]);

	return getMax();
}

int updateX(int pos, int val) {
    pos++;
    //updPref (pos, modInv(x[pos]));
    //if (x[pos] > 1) upd2(pos, -1);
    x[pos] = val;
    updPref (pos, x[pos]);
    //if (x[pos] > 1) upd2(pos, 1);
    upd2(pos, (x[pos] > 1));
	return getMax();
}

int updateY(int pos, int val) {
    pos++;
    y[pos] = val;
    updTree(pos, y[pos]);
	return getMax();
}


/*

3
2 1 3
3 4 1
1
2 1 2
ans: 8 6

10
2 1 1 1 1 2 1 1 1 1
4 5 6 1 4 5 6 1 100 1
0
ans: 400

*/

컴파일 시 표준 에러 (stderr) 메시지

horses.cpp: In function 'll getMax()':
horses.cpp:151:31: warning: conversion from 'll' {aka 'long long int'} to 'int' may change value [-Wconversion]
  151 |             int mxId = getMaxY(i, le);
      |                        ~~~~~~~^~~~~~~
horses.cpp: In function 'int init(int, int*, int*)':
horses.cpp:181:15: warning: conversion from 'll' {aka 'long long int'} to 'int' may change value [-Wconversion]
  181 |  return getMax();
      |         ~~~~~~^~
horses.cpp: In function 'int updateX(int, int)':
horses.cpp:192:15: warning: conversion from 'll' {aka 'long long int'} to 'int' may change value [-Wconversion]
  192 |  return getMax();
      |         ~~~~~~^~
horses.cpp: In function 'int updateY(int, int)':
horses.cpp:199:15: warning: conversion from 'll' {aka 'long long int'} to 'int' may change value [-Wconversion]
  199 |  return getMax();
      |         ~~~~~~^~

• Subtask 1: 17 / 17
• Subtask 2: 17 / 17
• Subtask 3: 20 / 20
• Subtask 4: 23 / 23
• Subtask 5: 0 / 23