数列分块入门 2

思路：先分块，然后每一块都用vector存起来，然后从小到大排序，查询的时候直接二分查找即可，坑点：不要在查找的时候排序，要每次改为就排序，否则会

超时，超了三次.........

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include <sstream>
#include<vector>
#include<cmath>    
#include<stack>
#include<random>
using namespace std;
#define io ios::sync_with_stdio(0),cin.tie(0)
#define ms(arr) memset(arr,0,sizeof(arr))
#define LD long double
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define inf 1<<30
#define  ull unsigned long long
const int Mod = 998244353;
const int maxn = 1e6 + 5;
int read()
{
    int x = 0, f = 1; char ch = getchar();
    while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}
int n, opt, l, r, c, block;
int belong[maxn], change[maxn];
int a[maxn];
vector<int>vt[5000];
void build()
{
    block = sqrt(n);//分块操作，每一个块的大小为sqrt(n)
    for (int i = 1; i <= n; i++) {        //遍历n个数
        belong[i] = (i - 1) / block + 1;//确定第i个数属于第几个块
        vt[belong[i]].push_back(a[i]);//进入队列
    }
    for (int i = 1; i <= belong[n]; i++) {//对每一块进行排序
        sort(vt[i].begin(), vt[i].end());
    }
}
void rest(int x)
{
    vt[x].clear();//清空这块
    for (int i = block * (x - 1) + 1; i <= min(block * x, n); i++) {//重新遍历这块
        vt[x].push_back(a[i]);
    }
    sort(vt[x].begin(), vt[x].end());//排序....坑点如果不在这里排序而是在查询的时候排序会因为查询次数过多而超时
}
void add(int x, int y, int val)
{
    for (int i = x; i <= min(block * belong[x], y); i++) {
        a[i] += val;
    }
    rest(belong[x]);//重排
    if (belong[x] != belong[y]) {//不在同一块
        for (int i = (belong[y] - 1) * block + 1; i <= y; i++) {//对y所在块遍历[左边界，y]
            a[i] += val;
        }
        rest(belong[y]);
        for (int i = belong[x] + 1; i < belong[y]; i++) {//遍历[x所在的块+1,y所在的块-1]，即中间块
            change[i] += val;
        }    
    }
}
void query(int x, int y, int k)
{
    LL ans = 0;
    for (int i = x; i <= min(block * belong[x], y); i++) {//遍历找符合条件的数
        if (a[i] + change[belong[x]] < k)ans++;
    }
    if (belong[x] != belong[y]) {//不在同一块
        for (int i = (block * (belong[y] - 1) + 1); i <= y; i++) {//找y块的
            if (a[i] + change[belong[y]] < k)ans++;
        }
        for (int i = belong[x] + 1; i < belong[y]; i++) {//找x,y中间块
            int t = k - change[i];
            //之前已经排序好了，不必再排
            ans += lower_bound(vt[i].begin(), vt[i].end(), t) - vt[i].begin();
        }
    }
    printf("%lld\n", ans);
}

int main() {
    io;
    n = read();
    for (int i = 1; i <= n; i++) {
        a[i] = read();
    }
    build();//分块
    for (int i = 1; i <= n; i++) {
        opt = read(); l = read(); r = read(); c = read();
        if (opt == 0) {
            add(l, r, c);
        }
        else {
            query(l, r, c * c);
        }
    }
    return 0;
}