CodeForces：Addition and First element at least X一起养成写作习惯！这是我

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题目描述

C - Segment Tree, part 2 - Addition and First element at least X

There is an array of n elements, initially filled with zeros. You need to write a data structure that processes two types of queries:

add v to all elements on the segment from l to r−1, find the minimum index j such that j≥l and a[j]≥x. Input The first line contains two numbers n and m (1≤n,m≤100000), the size of the array and the number of operations. The following lines contain the description of the operations. The description of each operation is as follows:

1 l r v: add v to all elements on the segment from l to r−1 (0≤i<n, 0≤v≤10^4). 2 x l: find the minimum index j such that j≥l and a[j]≥x (0≤x≤10^9, 0≤l<n). If there is no such element, print −1. Indices start from 0. Output For each operation of the second type, print the answer for the query.

Example

input

output

问题解析

题目是说给你一个初始为0的数组，进行两种操作，一个是把区间l~r的数都加上v，一种是让你找到下标大于等于j，且值大于等于x的第一个节点的位置。

区间加就是很基础的标记下传，这里主要讲第二种操作，线段树节点记录的是区间内的最大值，每次我们判断一下，如果左子节点的值大于x，那我们优先去左边找，如果找到的节点下标小于j，那我们就返回一个-1，然后回到上一层去右边找，也就是说我们这个线段树要有个回溯功能，如果再去右边找，找到的节点是小于x或是下标小于j的，那也返回-1。

但是这种做法，最坏情况下我们一次查询可能要把整个线段树跑一边，相当慢了，所以我们要减枝。当当前区间的最大值是小于x时，我们也不用纠结去左或右了，直接返回-1即可。或者当前区间的下标都是小于j的，我们也可直接返回-1。

#include<iostream>
using namespace std;
#include<vector>
#include<algorithm>
#include<math.h>
#include<set>
#include<numeric>
#include<string>
#include<string.h>
#include<iterator>
#include<map>
#include<unordered_map>
#include<stack>
#include<list>
#include<queue>
#include<iomanip>

#define endl '\n';
typedef long long ll;
typedef pair<ll, ll>PII;
const int N = 100500;
ll f[4 * N], b[4 * N], lmax[4 * N], rmax[4 * N], date[4 * N];

void push_down(ll k, ll l, ll r)
{
    if (b[k] != 0)
    {
        int m = (l + r) / 2;
        b[k + k] += b[k];
        b[k + k + 1] += b[k];
        f[k + k] += b[k];
        f[k + k + 1] += b[k];
        b[k] = 0;
    }
}

void revise(ll k, ll l, ll r, ll x, ll y, ll v)
{
    if (l == x && r == y)
    {
        b[k] += v;
        f[k] += v;
        return;
    }
    push_down(k, l, r);
    
    int m = (l + r) / 2;
    if (y <= m)revise(k + k, l, m, x, y, v);
    else
        if (x > m)revise(k + k + 1, m + 1, r, x, y, v);
        else
        {
            revise(k + k, l, m, x, m, v);
            revise(k + k + 1, m + 1, r, m + 1, y, v);
        }
    f[k] = max(f[k + k], f[k + k + 1]);
}

ll calc(int k, int l, int r, int x, int y)
{
    if (l == r)
    {
        if (l < y || f[k] < x)return -1;
        return l;
    }
    push_down(k, l, r);
    int m = (l + r) / 2;
    if (f[k] < x || r < y)return -1;
    int res = -1;
    if (f[k + k] >= x)res = calc(k + k, l, m, x, y);
    if (res == -1)res = calc(k + k + 1, m + 1, r, x, y);
    f[k] = max(f[k + k], f[k + k + 1]);
    return res;
}

int main()
{
    int n, m;
    cin >> n >> m;
    while (m--)
    {
        ll t, x, y, v;
        cin >> t >> x >> y;
        if (t == 1)
        {
            cin >> v;
            revise(1, 1, n, x + 1, y, v);
        }
        else cout << max((ll)-1, calc(1, 1, n, x, y + 1) - 1) << endl;
    }
    return 0;
}