uva12003 Array Transformer 分块

题目：

https://vjudge.net/problem/UVA-12003

题意：

给出一个长度为n的序列a，有m次操作，每次操作(L,R,v,p)，首先找出[L,R]内严格小于v的元素个数k，然后把a[p]更新为u∗k/(R−L+1)。输出最终的序列

思路：

无脑分块

#include <bits/stdc++.h>

using namespace std;

const int N = 300000 + 10;

int L[N], R[N], pos[N];
int sz, block;

int a[N], b[N];

void reset(int x)
{
    for(int i = L[x]; i <= R[x]; i++) b[i] = a[i];
    sort(b + L[x], b + 1 + R[x]);
}
void init(int n)
{
    block = (int)sqrt(n * log(n) / 2.0);
    //block = (int)sqrt(n);
    sz = n / block;
    if(n % block) sz++;
    for(int i = 1; i <= n; i++) pos[i] = (i-1) / block + 1;
    for(int i = 1; i <= sz; i++)
    {
        L[i] = (i-1) * block + 1;
        R[i] = i * block;
    }
    R[sz] = n;
    for(int i = 1; i <= sz; i++) reset(i);
}
int query(int l, int r, int x)
{
    int lb = pos[l], rb = pos[r];
    int ans = 0;
    if(lb == rb)
    {
        for(int i = l; i <= r; i++)
            if(a[i] < x) ans++;
    }
    else
    {
        for(int i = l; i <= R[lb]; i++)
            if(a[i] < x) ans++;
        for(int i = L[rb]; i <= r; i++)
            if(a[i] < x) ans++;
        for(int i = lb+1; i < rb; i++)
            ans += lower_bound(b + L[i], b + 1 + R[i], x) - (b + L[i]);
    }
    return ans;
}
void update(int x, int val)
{
    a[x] = val;
    reset(pos[x]);
}
int main()
{
    int n, m, u;
    while(~ scanf("%d%d%d", &n, &m, &u))
    {
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        init(n);
        int l, r, v, p;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%d%d", &l, &r, &v, &p);
            int k = query(l, r, v);
            update(p, 1LL * u * k / (r-l+1));
        }
        for(int i = 1; i <= n; i++) printf("%d\n", a[i]);
    }
    return 0;
}