제출 #1277502

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1277502		k12_khoi	말 (IOI15_horses)	C++17	34 / 100	1598 ms	57176 KiB

horses

#include "horses.h"

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

#define ll long long
#define pii pair<ll,ll>
#define X first
#define Y second

const int N = 500000 + 5;
const int mod = 1000000007;
const int limit = 1000000000;

int n, request, type, x, y, u, v, k;
int a[N], b[N], Xarr[N], Yarr[N], t[4 * N], lazy[4 * N], tTwo[4 * N];
set<int> se;
set<int>::iterator it;

int mul(int x, int y) {
    return int((1LL * x * y) % mod);
}

int luythua(int x, int nexp) {
    int c = 1;
    while (nexp) {
        if (nexp & 1) c = mul(c, x);
        x = mul(x, x);
        nexp >>= 1;
    }
    return c;
}

void down(int id) {
    if (lazy[id] != 1) {
        lazy[id << 1] = mul(lazy[id << 1], lazy[id]);
        lazy[id << 1 | 1] = mul(lazy[id << 1 | 1], lazy[id]);

        tTwo[id << 1] = mul(tTwo[id << 1], lazy[id]);
        tTwo[id << 1 | 1] = mul(tTwo[id << 1 | 1], lazy[id]);

        lazy[id] = 1;
    }
}

// update_range uses globals u,v,k like original
void update_range(int id, int l, int r) {
    if (u > r || v < l) return;
    if (u <= l && r <= v) {
        tTwo[id] = mul(tTwo[id], k);
        lazy[id] = mul(lazy[id], k);
        return;
    }
    down(id);
    int m = (l + r) >> 1;
    update_range(id << 1, l, m);
    update_range(id << 1 | 1, m + 1, r);
}

// update_point uses globals u,k: set t at leaf u to k
void update_point(int id, int l, int r) {
    if (l == r) {
        t[id] = k;
        return;
    }
    int m = (l + r) >> 1;
    if (u <= m) update_point(id << 1, l, m);
    else update_point(id << 1 | 1, m + 1, r);
    t[id] = max(t[id << 1], t[id << 1 | 1]);
}

// get_range uses globals u,v
int get_range(int id, int l, int r) {
    if (u > r || v < l) return 0;
    if (u <= l && r <= v) return t[id];
    down(id);
    int m = (l + r) >> 1;
    return max(get_range(id << 1, l, m), get_range(id << 1 | 1, m + 1, r));
}

// get_point uses global u
int get_point(int id, int l, int r) {
    if (l == r) return tTwo[id];
    down(id);
    int m = (l + r) >> 1;
    if (u <= m) return get_point(id << 1, l, m);
    else return get_point(id << 1 | 1, m + 1, r);
}

// build using pref_mod to avoid side effects of modifying cur inside recursion
function<void(int,int,int, const vector<int>&)> build; // forward

int init(int nn, int aarr[], int barr[]) {
    ::n = nn;
    for (int i = 0; i < n; ++i) {
        Xarr[i] = aarr[i];
        Yarr[i] = barr[i];
    }

    // prepare prefix modulo array (prefix_mod[i] = X[0]*...*X[i-1] mod)
    vector<int> prefix_mod(n + 1);
    prefix_mod[0] = 1;
    for (int i = 1; i <= n; ++i) prefix_mod[i] = mul(prefix_mod[i - 1], Xarr[i - 1]);

    // init segment tree arrays (clean)
    int SZ = 4 * (n + 5);
    for (int i = 0; i < SZ; ++i) {
        t[i] = 0;
        lazy[i] = 1;
        tTwo[i] = 1;
    }

    // build function (uses prefix_mod)
    build = [&](int id, int l, int r, const vector<int> &pref_mod) {
        lazy[id] = 1;
        tTwo[id] = 1;
        if (l == r) {
            if (l == 0) t[id] = 1;
            else t[id] = Yarr[l - 1];
            tTwo[id] = pref_mod[l]; // prefix mod for leaf l
            return;
        }
        int m = (l + r) >> 1;
        build(id << 1, l, m, pref_mod);
        build(id << 1 | 1, m + 1, r, pref_mod);
        t[id] = max(t[id << 1], t[id << 1 | 1]);
    };

    // build tree
    build(1, 0, n, prefix_mod);

    // fill set se
    se.clear();
    se.insert(0);
    for (int i = 0; i < n; ++i) if (Xarr[i] != 1) se.insert(i + 1);

    // find best: iterate as you requested: multiply cur by X[old-1] then --it then check pos=*it
    it = prev(se.end());
    int best = *it;
    // ma: pair(maxY, den) where den is cur-ref per your semantics; init with last element
    u = *it; v = n;
    pii ma = make_pair(get_range(1, 0, n), 1);
    long long cur = 1;

    while (it != se.begin()) {
        int old = *it;
        if (old) {
            cur = cur * 1LL * Xarr[old - 1];
            if (cur >= limit) cur = limit; // cap to avoid overflow
        }

        --it;
        int pos = *it;
        u = pos; v = n;
        int temp = get_range(1, 0, n);

        // compare ratios: ma.first/ma.second < temp/cur
        if ((long long)ma.first * cur < (long long)temp * ma.second) {
            best = pos;
            ma.first = temp;
            ma.second = 1; // follow your desired reset behavior
            cur = 1;
        }
    }

    u = best;
    int prefMod = get_point(1, 0, n);
    u = best; v = n;
    int ans = mul(prefMod, get_range(1, 0, n));
    return ans;
}

int updateX(int pos0, int val) {
    int pos = pos0 + 1;
    if (Xarr[pos - 1] != val) {
        if (Xarr[pos - 1] == 1) se.insert(pos);

        k = mul(luythua(Xarr[pos - 1], mod - 2), val); // multiplier modulo
        u = pos; v = n;
        update_range(1, 0, n);

        Xarr[pos - 1] = val;

        if (val == 1) se.erase(pos);
    }

    // find best again with same iteration logic
    it = prev(se.end());
    int best = *it;
    u = *it; v = n;
    pii ma = make_pair(get_range(1, 0, n), 1);
    long long cur = 1;

    while (it != se.begin()) {
        int old = *it;
        if (old) {
            cur = cur * 1LL * Xarr[old - 1];
            if (cur >= limit) cur = limit;
        }

        --it;
        int pos2 = *it;
        u = pos2; v = n;
        int temp = get_range(1, 0, n);

        if ((long long)ma.first * cur < (long long)temp * ma.second) {
            best = pos2;
            ma.first = temp;
            ma.second = 1;
            cur = 1;
        }
    }

    u = best;
    int prefMod = get_point(1, 0, n);
    u = best; v = n;
    return mul(prefMod, get_range(1, 0, n));
}

int updateY(int pos0, int val) {
    int pos = pos0 + 1;
    if (Yarr[pos - 1] != val) {
        u = pos; k = val;
        update_point(1, 0, n); // set t at leaf pos to val
        Yarr[pos - 1] = val;
    }

    // find best again
    it = prev(se.end());
    int best = *it;
    u = *it; v = n;
    pii ma = make_pair(get_range(1, 0, n), 1);
    long long cur = 1;

    while (it != se.begin()) {
        int old = *it;
        if (old) {
            cur = cur * 1LL * Xarr[old - 1];
            if (cur >= limit) cur = limit;
        }

        --it;
        int pos2 = *it;
        u = pos2; v = n;
        int temp = get_range(1, 0, n);

        if ((long long)ma.first * cur < (long long)temp * ma.second) {
            best = pos2;
            ma.first = temp;
            ma.second = 1;
            cur = 1;
        }
    }

    u = best;
    int prefMod = get_point(1, 0, n);
    u = best; v = n;
    return mul(prefMod, get_range(1, 0, n));
}

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