LibreOJ #6279

题目链接:#6279. 数列分块入门 3

题目大意

给出一个长为 \(n\) 的数列，以及 \(n\) 个操作，操作涉及区间加法，询问区间内小于某个值 \(x\) 的前驱（比其小的最大元素）。

solution

这就是标准的分块套平衡树的裸题

但是平衡树太难写了,这里我们就对每个块排个序就好了

我们还是沿用第二题的写法

然后,查询就变成了找最大值,同时记录一下个数,如果这个个数为0, 那就是-1, 否则就是我们找的值

注意:在 \(lower_bound\) 时, 最后的最大值别忘了加上我们记录的值哟

Code:

/**
*    Author: Alieme
*    Data: 2020.9.8
*    Problem: LibreOJ #6279
*    Time: O()
*/
#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>

#define int long long
#define rr register

#define inf 1e12
#define MAXN 100010

using namespace std;

inline int read() {
	int s = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

void print(int x) {
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) print(x / 10);
	putchar(x % 10 + 48);
}

int n, len;

int a[MAXN], b[MAXN], id[MAXN], v[MAXN];

inline void update(int x) {
	int l = (x - 1) * len + 1, r = min(x * len, n);
	for (rr int i = l; i <= r; i++) b[i] = a[i];
	sort(b + l, b + r + 1);
}

inline void add(int l, int r, int x) {
	int start = id[l], end = id[r];
	if (start == end) {
		for (rr int i = l; i <= r; i++) a[i] += x;
		update(id[l]);
		return ;
	}
	for (rr int i = l; id[i] == start; i++) a[i] += x; update(start);
	for (rr int i = start + 1; i < end; i++) v[i] += x;
	for (rr int i = r; id[i] == end; i--) a[i] += x; update(end);
}

inline int fin(int pid, int x, int &ans) {
	int l = (pid - 1) * len + 1, r = min(pid * len, n);
	int t = lower_bound(b + l, b + r + 1, x - v[pid]) - (b + l);
	ans += t;
	// cout << t;
	// cout << t << " " << b[t + l - 1] << " "  << x - v[pid] << "\n";
	if (t == 0) return -inf;
	return b[t - 1 + l] + v[pid];
}

inline int query(int l, int r, int x) {
	int maxx = -inf, ans = 0, start = id[l], end = id[r];
	if (start == end) {
		for (rr int i = l; i <= r; i++) if (a[i] + v[start] < x) maxx = max(maxx, a[i] + v[start]), ans++;
		if (ans == 0) maxx = -1;
		return maxx;
	}
	for (rr int i = l; id[i] == start; i++) if (a[i] + v[start] < x) maxx = max(maxx, a[i] + v[start]), ans++;
	for (rr int i = start + 1; i < end; i++) maxx = max(maxx, fin(i, x, ans));
	for (rr int i = r; id[i] == end; i--) if (a[i] + v[end] < x) maxx = max(maxx, a[i] + v[end]), ans++;
	if (ans == 0) maxx = -1;
	return maxx;
}

signed main() {
	// freopen("a1.in", "r", stdin);
	// freopen("a.out", "w", stdout);
	n = read(); 
	len = sqrt(n);
	for (rr int i = 1; i <= n; i++) a[i] = read(), id[i] = (i - 1) / len + 1;
	for (rr int i = len; i <= n; i += len) update(id[i]);
	for (rr int i = 1; i <= n; i++) {
		int opt = read(), l = read(), r = read(), c = read();
		if (opt == 0) add(l, r, c);
		if (opt == 1) cout << query(l, r, c) << "\n";
	}
}