Find the median (2019 Nian cattle off more school seventh title + E left and right to open and close tree line)

Topic Link

Portal

The meaning of problems

Each set which adds to the number of period of continuous section and then query the median in the current set.

Thinking

I want good ideas, but the memory card.

Then write a prescription dynamic segment tree card did not last, game room with big brother had a dynamic prescription, \ (TQL \) .

But opinions on the card can only add discrete left and right opening and closing segment tree wrote.

Code

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson (rt<<1),L,mid
#define rson (rt<<1|1),mid + 1,R
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 800000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int n, tot, x, y, xx, yy, m1, m2, a1, a2, b1, b2, c1, c2;
int L[maxn], R[maxn], num[maxn*2], lazy[maxn*4];
LL sum[maxn*4];

void push_up(int rt) {
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}

void push_down(int rt, int l, int r) {
    if(!lazy[rt]) return;
    int x = lazy[rt];
    lazy[rt] = 0;
    int mid = (l + r) >> 1;
    lazy[rt<<1] += x, lazy[rt<<1|1] += x;
    sum[rt<<1] += 1LL * x * (num[mid+1]-num[l]);
    sum[rt<<1|1] += 1LL * x * (num[r+1]-num[mid+1]);
}

void update(int l, int r, int rt, int L, int R) {
    if(l <= L && R <= r) {
        sum[rt] += num[R+1] - num[L];
        ++lazy[rt];
        return;
    }
    push_down(rt, L, R);
    int mid = (L + R) >> 1;
    if(r <= mid) update(l, r, lson);
    else if(l > mid) update(l, r, rson);
    else {
        update(l, mid, lson);
        update(mid + 1, r, rson);
    }
    push_up(rt);
}

int query(LL all, int rt, int L, int R) {
    if(L == R) {
        LL pos = sum[rt] / (num[R+1] - num[L]);
        pos = num[L] + (all - 1) / pos;
        return pos;
    }
    push_down(rt, L, R);
    int mid = (L + R) >> 1;
    if(sum[rt<<1] >= all) return query(all, lson);
    else return query(all - sum[rt<<1], rson);
}

int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif
    scanf("%d", &n);
    scanf("%d%d%d%d%d%d", &x, &xx, &a1, &b1, &c1, &m1);
    scanf("%d%d%d%d%d%d", &y, &yy, &a2, &b2, &c2, &m2);
    L[1] = min(x, y) + 1, R[1] = max(x, y) + 1;
    L[2] = min(xx, yy) + 1, R[2] = max(xx, yy) + 1;
    num[++tot] = L[1], num[++tot] = R[1] + 1;
    num[++tot] = L[2], num[++tot] = R[2] + 1;
    for(int i = 3; i <= n; ++i) {
        int num1 = ((1LL * a1 * xx % m1 + 1LL * b1 * x % m1) % m1 + c1) % m1;
        int num2 = ((1LL * a2 * yy % m2 + 1LL * b2 * y % m2) % m2 + c2) % m2;
        L[i] = min(num1, num2) + 1, R[i] = max(num1, num2) + 1;
        x = xx, xx = num1;
        y = yy, yy = num2;
        num[++tot] = L[i], num[++tot] = R[i] + 1;
    }
    sort(num + 1, num + tot + 1);
    tot = unique(num + 1, num + tot + 1) - num - 1;
    LL all = 0;
    for(int i = 1; i <= n; ++i) {
        all += R[i] - L[i] + 1;
        L[i] = lower_bound(num + 1, num + tot + 1, L[i]) - num;
        R[i] = lower_bound(num + 1, num + tot + 1, R[i] + 1) - num;
        update(L[i], R[i]-1, 1, 1, tot);
        printf("%d\n", query((all + 1) /2, 1, 1, tot));
    }
    return 0;
}