table of Contents
@ (2019 seventh cattle off E_Find the median closing left and right open segment tree)
The meaning of problems
Link: here Wallpaper
I understand the meaning of the questions is: initial sequence is empty, there \ (n (400000) \) operations, each operation the interval \ ([Li, Ri] \ ) is added to the digital sequence, there is a sequence automatically sequence, the output operation after each median much.
Want to feel when the game method should also be able to write, it makes sense that trouble may be, the answer is probably half again messing about. .
An analytical
A routine: left and right to open and close the segment tree
or tree line like weights, but the value is not a point of leaf node represents, but rather the representation of the range of the interval.
Added \ (n \) a range intervals, they will cut this large range of many parts. Clearly adding \ ([Li, Ri] \ ) when the range, can be decomposed into a number of segments added range interval.
Every point, as a right-left opening and closing section, all the \ (Li, Ri + 1 \ ) discretization down.
Then update operation is to do a range of addition, a one point increase representing the interval it represents each value appears once.
And the general query tree line weights the same query. Leaf node is found, first calculate the number of times each value occurs in this interval \ (len \) , known I was looking for in the first row (p \) \ digit, the left end point of this range represents the point is \ ( L \) , the answer is: \ ((the p-+ len-1) / len + L-1 \) .
AC_Code
#pragma comment(linker, "/STACK:102400000,102400000")
//#include<bits/stdc++.h>
#include <ctime>
#include <iostream>
#include <assert.h>
#include <vector>
#include <queue>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define fi first
#define se second
#define endl '\n'
#define o2(x) (x)*(x)
#define BASE_MAX 31
#define mk make_pair
#define eb push_back
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define clr(a, b) memset((a),(b),sizeof((a)))
#define iis std::ios::sync_with_stdio(false); cin.tie(0)
#define my_unique(x) sort(all(x)),x.erase(unique(all(x)),x.end())
using namespace std;
#pragma optimize("-O3")
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> pii;
inline LL read() {
LL x = 0;int f = 0;
char ch = getchar();
while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x = f ? -x : x;
}
inline void write(LL x, bool f) {
if (x == 0) {putchar('0'); if(f)putchar('\n');else putchar(' ');return;}
if (x < 0) {putchar('-');x = -x;}
static char s[23];
int l = 0;
while (x != 0)s[l++] = x % 10 + 48, x /= 10;
while (l)putchar(s[--l]);
if(f)putchar('\n');else putchar(' ');
}
int lowbit(int x) { return x & (-x); }
template<class T>T big(const T &a1, const T &a2) { return a1 > a2 ? a1 : a2; }
template<class T>T sml(const T &a1, const T &a2) { return a1 < a2 ? a1 : a2; }
template<typename T, typename ...R>T big(const T &f, const R &...r) { return big(f, big(r...)); }
template<typename T, typename ...R>T sml(const T &f, const R &...r) { return sml(f, sml(r...)); }
void debug_out() { cerr << '\n'; }
template<typename T, typename ...R>void debug_out(const T &f, const R &...r) {cerr << f << " ";debug_out(r...);}
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]: ", debug_out(__VA_ARGS__);
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int HMOD[] = {1000000009, 1004535809};
const LL BASE[] = {1572872831, 1971536491};
const int mod = 1e9 + 0;//998244353
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MXN = 1e6 + 7;
const int MXE = 1e6 + 7;
int n, m;
int xs[MXN], ys[MXN], ls[MXN], rs[MXN];
LL a1, a2, b1, b2, c1, c2, m1, m2;
vector<int> vs;
int lazy[MXN<<2];
LL sum[MXN<<2];
void push_down(int rt, int l, int mid, int r) {
if(lazy[rt] == 0) return;
lazy[rt<<1] += lazy[rt], lazy[rt<<1|1] += lazy[rt];
sum[rt<<1] += (LL)lazy[rt] * (vs[mid + 1] - 1 - (vs[l] - 1));
sum[rt<<1|1] += (LL)lazy[rt] * (vs[r + 1] - 1 - (vs[mid + 1] - 1));
lazy[rt] = 0;
}
void update(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
sum[rt] += vs[R + 1] - 1 - (vs[L] - 1);
// debug(L, R, vs[R+1] - 1, vs[L] - 1)
++ lazy[rt];
return;
}
int mid = (l + r) >> 1;
push_down(rt, l, mid, r);
if(L > mid) update(L, R, mid + 1, r, rt<<1|1);
else if(R <= mid) update(L, R, l, mid, rt<<1);
else {
update(L, mid, l, mid, rt<<1), update(mid + 1, R, mid + 1, r, rt<<1|1);
}
sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
int query(LL p, int l, int r, int rt) {
if(l == r) {
// debug(l, vs[l], p, sum[rt])
LL len = sum[rt]/(vs[l+1] - vs[l]);
return (p + len - 1)/len + vs[l] - 1;
}
int mid = (l + r) >> 1;
push_down(rt, l, mid, r);
// debug(rt, sum[rt], sum[rt<<1], sum[rt<<1|1])
if(sum[rt<<1] >= p) return query(p, l, mid, rt<<1);
else return query(p - sum[rt<<1], mid + 1, r, rt<<1|1);
}
int main() {
#ifndef ONLINE_JUDGE
freopen("/home/cwolf9/CLionProjects/ccc/in.txt", "r", stdin);
// freopen("/home/cwolf9/CLionProjects/ccc/out.txt", "w", stdout);
#endif
n = read();
xs[1] = read(), xs[2] = read(), a1 = read(), b1 = read(), c1 = read(), m1 = read();
ys[1] = read(), ys[2] = read(), a2 = read(), b2 = read(), c2 = read(), m2 = read();
vs.eb(0);
vs.eb(ls[1] = sml(xs[1], ys[1]) + 1), vs.eb((rs[1] = big(xs[1], ys[1]) + 1)+1);
vs.eb(ls[2] = sml(xs[2], ys[2]) + 1), vs.eb((rs[2] = big(xs[2], ys[2]) + 1)+1);
for(int i = 3; i <= n; ++i) {
xs[i] = (xs[i-1] * a1 + xs[i-2] * b1 + c1)%m1, ys[i] = (ys[i-1] * a2 + ys[i-2] * b2 + c2)%m2;
vs.eb(ls[i] = sml(xs[i], ys[i]) + 1), vs.eb((rs[i] = big(xs[i], ys[i]) + 1)+1);
}
my_unique(vs);
for(auto x: vs) printf("%d ", x); printf("\n");
for(int i = 1, tx, ty; i <= n; ++i) {
tx = lower_bound(all(vs), ls[i]) - vs.begin();
ty = upper_bound(all(vs), rs[i]) - vs.begin();
// debug(tx, ty, vs.size())
update(tx, ty - 1 , 1, vs.size(), 1);
// debug(sum[1], sum[1]/2+(sum[1]%2))
printf("%d\n", query(sum[1]/2+(sum[1]%2), 1, vs.size(), 1));
// debug(sum[1])
// if(i == 2) break;
}
return 0;
}