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P13977数列分块入门 2参与者 2已保存回复 3

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@mlje2xv2
此快照首次捕获于
2026/02/12 19:42
7 天前
此快照最后确认于
2026/02/15 11:20
4 天前
查看原帖
CPP
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 200005;
int n, a[N], b[N], L[N], R[N], bl[N], lzy[N], block;
void add(int l, int r, int x)
{
    int lb = bl[l], rb = bl[r];
    if(lb == rb)
    {
        for(int i = l; i <= r; i ++)
        {
            a[i] += x;
            b[i] = a[i];
        }
        sort(b + L[lb], b + R[rb] + 1);
        return ;
    }
    for(int i = l; i <= R[lb]; i ++)
    {
        a[i] += x;
        b[i] = a[i];
    }
    for(int i = L[rb]; i <= r; i ++)
    {
        a[i] += x;
        b[i] = a[i];
    }
    sort(b + L[lb], b + R[lb] + 1);
    sort(b + L[rb], b + R[rb] + 1);
    for(int i = lb + 1; i <= rb - 1; i ++)
    {
        lzy[i] += x;
    }   
}
int query(int l, int r, int x)
{
    int ans = 0, lb = bl[l], rb = bl[r];
    if(lb == rb)
    {
        for(int i = l; i <= r; i ++)
        {
            if(a[i] + lzy[lb] < x)
            {
                ans ++;
            }
        }
        return ans;
    }
    for(int i = l; i <= R[lb]; i ++)
    {
        ans += (a[i] + lzy[lb] < x);
    }
    for(int i = L[rb]; i <= r; i ++)
    {
        ans += (a[i] + lzy[rb] < x);
    }
    for(int i = lb + 1; i <= rb - 1; i ++)
    {
        ans += (lower_bound(b + L[i], b + R[i] + 1, x - lzy[i]) - b - L[i]);
    }
    return ans;
}
signed main()
{
    cin >> n;
    block = sqrt(n);    
    for(int i = 1; i <= n; i ++)
    {
        cin >> a[i];
        b[i] = a[i];
    }
    for(int i = 1; i <= block; i ++)
    {
        L[i] = n / block * (i - 1) + 1;
        R[i] = n / block * i;
    }
    R[block] = n;
    for(int i = 1; i <= block; i ++)
    {
        for(int j = L[i]; j <= R[i]; j ++)
        {
            bl[j] = i;
        }
        sort(b + L[i], b + R[i] + 1);
    }
    for(int _ = 1; _ <= n; _ ++)
    {
        int op, l, r, c;
        cin >> op >> l >> r >> c;
        if(op == 0)
        {
            add(l, r, c);
            // for(int i = 1; i <= n; i ++)
            // {
            //     cout << a[i] + lzy[i] << " ";
            // }
            // cout << endl;
        }
        if(op == 1)
        {
            cout << query(l, r, c * c) << endl;
        }
    }
}

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