【线段树区间更新】 Naive Operations HDU6315 2018多校第二场

Naive Operations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)
Total Submission(s): 1977    Accepted Submission(s): 853

 

Problem Description

In a galaxy far, far away, there are two integer sequence a and b of length n.
b is a static permutation of 1 to n. Initially a is filled with zeroes.
There are two kind of operations:
1. add l r: add one for al,al+1...ar
2. query l r: query ∑ri=l⌊ai/bi⌋

 

 

Input

There are multiple test cases, please read till the end of input file.
For each test case, in the first line, two integers n,q, representing the length of a,b and the number of queries.
In the second line, n integers separated by spaces, representing permutation b.
In the following q lines, each line is either in the form 'add l r' or 'query l r', representing an operation.
1≤n,q≤100000, 1≤l≤r≤n, there're no more than 5 test cases.

 

 

Output

Output the answer for each 'query', each one line.

 

 

Sample Input

5 12

1 5 2 4 3

add 1 4

query 1 4

add 2 5

query 2 5

add 3 5

query 1 5

add 2 4

query 1 4

add 2 5

query 2 5

add 2 2

query 1 5

 

 

Sample Output

1 1 2 4 4 6

 

 

Source

2018 Multi-University Training Contest 2

 

 

/// 每个节点保存amax及bmin 若amax < bmin 无法更新
/// 若 amax >= bmin 不一定是否更新
/// lazy标记保存区间 + 1 次数

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

const int mn = 1e5 + 10;

int b[mn];

int id[mn];
struct node
{
	int l, r;
	int amax, bmin;
	int lazy, sum;
} t[8 * mn];

void pushup(int i)
{
	t[i].amax = max(t[2 * i].amax, t[2 * i + 1].amax);
	t[i].bmin = min(t[2 * i].bmin, t[2 * i + 1].bmin);
	t[i].sum = t[2 * i].sum + t[2 * i + 1].sum;
}

void build(int l, int r, int i)
{
	t[i].l = l, t[i].r = r;
	t[i].lazy = 0;
    if (l == r)
	{
		id[l] = i;
		t[i].sum = 0;
		t[i].amax = 0;
		t[i].bmin = b[l];
		return;
	}
	int m = (l + r) / 2;
	build(l, m, 2 * i);
	build(m + 1, r, 2 * i + 1);
	pushup(i);
	return;
}

void pushdown(int i)
{
	t[2 * i].lazy += t[i].lazy;
	t[2 * i + 1].lazy += t[i].lazy;
	t[2 * i].amax += t[i].lazy;
	t[2 * i + 1].amax += t[i].lazy;
	t[i].lazy = 0;
}

void update(int x, int y, int i)
{
	int l = t[i].l, r = t[i].r;
	int m = (l + r) / 2;

	if (x <= l && y >= r)  /// 区间包含时更新
	{
		t[i].amax++;
		if (t[i].amax < t[i].bmin)
		{
			t[i].lazy++;
			return;
		}
	}

	if (l == r)
	{
		if (t[i].amax >= t[i].bmin)
		{
			t[i].sum++;
			t[i].bmin += b[l];
		}
		return;
	}

	if (t[i].lazy > 0)
		pushdown(i);

	if (y <= m)
		update(x, y, 2 * i);
	else if (x > m)
		update(x, y, 2 * i + 1);
	else
	{
		update(x, m, 2 * i);
		update(m + 1, y, 2 * i + 1);
	}
	pushup(i);
	return;
}

int query(int x, int y, int i)
{
	int l = t[i].l, r = t[i].r;
	int m = (l + r) / 2;

	if (x == l && r == y)
		return t[i].sum;

	if (t[i].lazy > 0)
		pushdown(i);

    if (y <= m)
		return query(x, y, 2 * i);
	else if (x > m)
		return query(x, y, 2 * i + 1);
	else
		return query(x, m, 2 * i) + query(m + 1, y, 2 * i + 1);
}

int main()
{
	int n, q;
	while (~scanf("%d %d", &n, &q))
	{
		for (int i = 1; i <= n; i++)
			scanf("%d", &b[i]);

		build(1, n, 1);
        while (q--)
		{
			char ch[7];
			int x, y;
			scanf("%s %d %d", ch, &x, &y);
			if (ch[0] == 'a')
				update(x, y, 1);
			else if (ch[0] == 'q')
				printf("%d\n", query(x, y, 1));
		}
	}
	return 0;
}