제출 #171070

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
171070		hentai_lover	Segments (IZhO18_segments)	C++14	100 / 100	4997 ms	15632 KiB

segments

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define lft(x) x * 2
#define rgt(x) x * 2 + 1

#define tm hui_pizda
#define ft first
#define sd second
#define pb push_back
#define pf push_front
#define sz size()
#define cnt continue
#define m_p make_pair
#define fr(i, l, r) for(int i = l; i <= r; ++ i)
#define rf(i, r, l) for(int i = r; i >= l; -- i)
#define all(x) x.begin(), x.end()
//#pragma GCC optimize(-O3)
//#pragma GCC optimize(Ofast)
//#pragma GCC optimize("unroll-loops")

using namespace __gnu_pbds;
using namespace std;

template <typename T>
using _set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef int ll;
typedef long double ld;

typedef pair <ll, ll> pll;
typedef pair <int, int> pii;
typedef vector <int> vi;
typedef vector <ll> vl;
typedef vector <pii> vpi;
typedef vector <pll> vpl;

mt19937_64 rnd(time(NULL));

const ll N = 1e6;
const ll mtrxN = 10;
const ll oo = 1e18 + 10;
const ll B = 500;
const ll mod = 1e9 + 7;
const ll B1 = 1500;
const ll B2 = 1500;


struct mtrx{
    ll m[mtrxN][mtrxN] = {};
};
mtrx mtrx_mult(mtrx a, mtrx b){
    mtrx c;
    fr(i, 0, mtrxN - 1){
        fr(j, 0, mtrxN - 1){
            ll sum = 0;
            fr(x, 0, mtrxN - 1){
                sum += a.m[i][x] * b.m[x][j];
                sum %= mod;
            }
            c.m[i][j] = sum;
        }
    }
    return c;
}
mtrx mtrx_pow(mtrx a, ll n){
    mtrx res;
    fr(i, 0, mtrxN - 1)fr(j, 0, mtrxN - 1)res.m[i][j] = a.m[i][j];
    n --;
    while(n){
        if(n&1)res = mtrx_mult(res, a);
        a = mtrx_mult(a, a);
        n >>= 1;
    }
    return res;
}
ll _pow(ll a, ll n){
    ll r = 1;
    while(n){
        if(n&1)r = r * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return r;
}
ll div(ll x, ll y, ll md){
    return x * _pow(y, md - 2) % md;
}

ll lastans, last = 1, n, t, i, a, b, k, l, r, now, usd[N], mn[N], mx[N];
vpl temp, sqd1, sqd2, deleted, aded;
vl sqd3, sqd4;
pll o[N];

ll cnv(ll a){
    return (t * lastans) ^ a;
}

ll get1(ll l, ll r, ll x){
    ll last = 0, ans = 0;
    for(ll i = B2; i < (ll)sqd1.sz; i += B2){
        if(mx[i] < l || mn[i] > r);
        else if(mn[i] < l && l <= mx[i] || mn[i] <= r && r < mx[i]){
            //посчитать вручную
            //cout << "last: " << last << " i: " << i << endl;
            fr(j, last, i - 1)
                if(sqd1[j].sd >= l && sqd1[j].sd <= r && sqd1[j].ft >= x)ans ++;
        }   else{
            ll pos = lower_bound(sqd3.begin() + last, sqd3.begin() + i, x) - sqd3.begin();
            ans += i - pos;
        }
        last = i;
    }
    fr(i, last, (ll)sqd1.sz - 1)
        if(sqd1[i].sd >= l && sqd1[i].sd <= r && sqd1[i].ft >= x)ans ++;
    return ans;
}

ll get2(ll l, ll r, ll x){
    ll last = 0, ans = 0;
    for(ll i = B2; i < (ll)sqd2.sz; i += B2){
        if(mx[i] < l || mn[i] > r);
        else if(mn[i] < l && l <= mx[i] || mn[i] <= r && r < mx[i]){
            //посчитать вручную
            fr(j, last, i - 1)
                if(sqd2[j].sd >= l && sqd2[j].sd <= r && sqd2[j].ft >= x)ans ++;
        }   else{
            ll pos = lower_bound(sqd4.begin() + last, sqd4.begin() + i, x) - sqd4.begin();
            ans += i - pos;
        }
        last = i;
    }
    fr(i, last, (ll)sqd2.sz - 1)
        if(sqd2[i].sd >= l && sqd2[i].sd <= r && sqd2[i].ft >= x)ans ++;
    return ans;
}

int main(){
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> t;
    fr(tr, 1, n){
        cin >> i;
        if(i == 1){
            cin >> a >> b;
            l = cnv(a);
            r = cnv(b);
            if(l > r)swap(l, r);
            o[++ now] = {l, r};
            aded.pb({l, r});
        }
        if(i == 2){
            ll d;
            cin >> d;
            //удалить отрезок l..r
            usd[d] = 1;
            deleted.pb({o[d].ft, o[d].sd});
        }

        if(i == 3){
            cin >> a >> b >> k;
            l = cnv(a);
            r = cnv(b);
            if(l > r)swap(l, r);

            //(кол-во чисел с li 1..l - 1 и ri >= l + k - 1) + (кол-во чисел с li l..r - k + 1 и ri - li + 1)

            //пройтись по накопленному вектору отрезков

            lastans = 0;
            for(auto i : aded)
                if(i.ft < l && i.sd >= l + k - 1 || i.ft >= l && i.ft <= r - k + 1 && i.sd - i.ft + 1 >= k)lastans ++;
            for(auto i : deleted)
                if(i.ft < l && i.sd >= l + k - 1 || i.ft >= l && i.ft <= r - k + 1 && i.sd - i.ft + 1 >= k)lastans --;


            //взять ответ из посчитанной ранее sqrtdecompompozition
            lastans += get1(1, l - 1, l + k - 1);
            lastans += get2(l, r - k + 1, k);
//            cout << "                 ";
            cout << lastans << "\n";
        }

        if(tr % B1 == 0){
            temp.clear();
            fr(i, 1, now){
                if(usd[i])cnt;
                temp.pb(o[i]);
            }
            sort(temp.begin(), temp.end());
            vpl temp2, temp3;
            aded.clear();
            deleted.clear();
            sqd1.clear();
            sqd2.clear();
            sqd3.clear();
            sqd4.clear();

            ll n = oo, x = -oo;
            fr(i, 0, (ll)temp.sz - 1){

                if(i != 0 && i % B2 == 0){
                    mn[i] = n, mx[i] = x;
                    sort(temp2.begin(), temp2.end());
                    sort(temp3.begin(), temp3.end());
                    for(auto i : temp2)sqd1.pb(i), sqd3.pb(i.ft);
                    for(auto i : temp3)sqd2.pb(i), sqd4.pb(i.ft);
                    n = oo, x = -oo;
                    temp2.clear();
                    temp3.clear();
                }
                temp2.pb({temp[i].sd, temp[i].ft});
                temp3.pb({temp[i].sd - temp[i].ft + 1, temp[i].ft});
                n = min(n, temp[i].ft);
                x = max(x, temp[i].ft);
            }
            sort(temp2.begin(), temp2.end());
            sort(temp3.begin(), temp3.end());
            for(auto i : temp2)sqd1.pb(i), sqd3.pb(i.ft);
            for(auto i : temp3)sqd2.pb(i), sqd4.pb(i.ft);

            last = now + 1;
        }
    }
    return 0;
}
/*

*/

컴파일 시 표준 에러 (stderr) 메시지

segments.cpp:42:20: warning: overflow in implicit constant conversion [-Woverflow]
 const ll oo = 1e18 + 10;
               ~~~~~^~~~
segments.cpp: In function 'll get1(ll, ll, ll)':
segments.cpp:103:27: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
         else if(mn[i] < l && l <= mx[i] || mn[i] <= r && r < mx[i]){
                 ~~~~~~~~~~^~~~~~~~~~~~~
segments.cpp: In function 'll get2(ll, ll, ll)':
segments.cpp:123:27: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
         else if(mn[i] < l && l <= mx[i] || mn[i] <= r && r < mx[i]){
                 ~~~~~~~~~~^~~~~~~~~~~~~
segments.cpp: In function 'int main()':
segments.cpp:171:29: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
                 if(i.ft < l && i.sd >= l + k - 1 || i.ft >= l && i.ft <= r - k + 1 && i.sd - i.ft + 1 >= k)lastans ++;
                    ~~~~~~~~~^~~~~~~~~~~~~~~~~~~~
segments.cpp:173:29: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
                 if(i.ft < l && i.sd >= l + k - 1 || i.ft >= l && i.ft <= r - k + 1 && i.sd - i.ft + 1 >= k)lastans --;
                    ~~~~~~~~~^~~~~~~~~~~~~~~~~~~~

• Subtask 1: 7 / 7
• Subtask 2: 15 / 15
• Subtask 3: 16 / 16
• Subtask 4: 17 / 17
• Subtask 5: 20 / 20
• Subtask 6: 25 / 25