分块 （区间加法，寻找区间比k小的数的个数

题目链接https://loj.ac/problem/6278

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
const int inf = 0x3f3f3f3f;
int dir[10][10] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int a[maxn], pos[maxn], opt, l, r, c, t, n, lz[maxn], ans;
vector<int> v[600];

inline int read()
{
    char ch = getchar(); int k = 0, f = 1;
    while(ch < '0' || ch > '9') {if(ch == '-')f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9') {k = k * 10 + ch - '0'; ch = getchar();}
    return k * f;
}

inline void reset(int x)//由于加法更新了值 要更新vector数组
{
    v[x].clear();
    for(int i = (x - 1) * t + 1; i <= min(n, x * t); ++i)
    {
        v[x].push_back(a[i]);
    }
    sort(v[x].begin(), v[x].end());//块中数从小到大
}

inline void inser(int l, int r, int c)
{
    for(int i = l; i <= min(r, pos[l] * t); ++i) a[i] += c;
    reset(pos[l]);
    if(pos[l] != pos[r])
    {
        for(int i = (pos[r] - 1) * t + 1; i <= r; ++i) a[i] += c;
        reset(pos[r]);
    }
    for(int i = pos[l] + 1; i <= pos[r] - 1; ++i) lz[i] += c;
}

inline int findy(int l, int r, int c)
{
    int ans = 0;
    for(int i = l; i <= min(r, pos[l] * t); ++i)
    {
        if(a[i] + lz[pos[l]] < c) ans++;
    }
    if(pos[l] != pos[r])
    {
        for(int i = (pos[r] - 1) * t + 1; i <= r; ++i)
        {
            if(a[i] + lz[pos[r]] < c) ans++;
        }
    }

    for(int i = pos[l] + 1; i <= pos[r] - 1; ++i)
    {
        ans += lower_bound(v[i].begin(), v[i].end(), c - lz[i]) - v[i].begin();
    }
    return ans;
}

int main()
{
    n = read();
    for(int i = 1; i <= n; ++i) cin>>a[i];
    t = sqrt(n);
    for(int i = 1; i <= n; ++i)
    {
        pos[i] = (i - 1) / t + 1;
        v[pos[i]].push_back(a[i]);
    }
    for(int i = 1; i <= pos[n]; ++i)
    {
        sort(v[i].begin(), v[i].end());
    }

    for(int i = 1; i <= n; ++i)
    {
        opt = read(); l = read(); r = read(); c = read();
        if(opt == 0) inser(l, r, c);
        else cout<<findy(l, r, c * c)<<endl;
    }
    return 0;
}