LOJ-数列分块入门5

链接:

https://loj.ac/problem/6281

题意:

给出一个长为 的数列 ，以及 n个操作，操作涉及区间开方，区间求和。

思路:

考虑开方5次之后就为1, 即考虑一整个区间的开方次数,对小于5次的区间暴力开方,否则就不管他.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 1e5+10;

LL a[MAXN], Tag[MAXN], Sum[MAXN];
LL Belong[MAXN];
int n, part;

void Update(LL l, LL r, LL c)
{
    if (Tag[Belong[l]] < 5)
        for (int i = l;i <= min(Belong[l]*part, r);i++)
        {
            Sum[Belong[i]] -= a[i];
            a[i] = sqrt(a[i]);
            Sum[Belong[i]] += a[i];
        }
    if (Belong[l] != Belong[r])
    {
        if (Tag[Belong[r]] < 5)
            for (int i = max((Belong[r]-1)*part+1, l);i <= r;i++)
            {
                Sum[Belong[i]] -= a[i];
                a[i] = sqrt(a[i]);
                Sum[Belong[i]] += a[i];
            }
    }
    for (int i = Belong[l]+1;i <= Belong[r]-1;i++)
    {
        if (Tag[i] < 5)
        {
            Tag[i]++;
            for (int j = (i-1)*part+1;j <= i*part;j++)
            {
                Sum[i] -= a[j];
                a[j] = sqrt(a[j]);
                Sum[i] += a[j];
            }
        }
    }
}

LL Query(LL l, LL r, LL c)
{
    LL res = 0;
    for (int i = l;i <= min(Belong[l]*part, r);i++)
        res += a[i];
    if (Belong[l] != Belong[r])
    {
        for (int i = max((Belong[r]-1)*part+1, l);i <= r;i++)
            res += a[i];
    }
    for (int i = Belong[l]+1;i <= Belong[r]-1;i++)
        res += Sum[i];
    return res;
}

int main()
{
    scanf("%d", &n);
    part = sqrt(n);
    for (int i = 1;i <= n;i++)
    {
        scanf("%lld", &a[i]);
        Belong[i] = (i-1)/part + 1;
        Sum[Belong[i]] += a[i];
    }
    int op, l, r, c;
    for (int i = 1;i <= n;i++)
    {
        scanf("%d", &op);
        if (op == 0)
        {
            scanf("%d%d%d", &l, &r, &c);
            Update(l, r, c);
        }
        else
        {
            scanf("%d%d%d", &l, &r, &c);
            printf("%lld\n", Query(l, r, c));
        }
    }

    return 0;
}