Loj 6281. 数列分块入门 5

链接：

https://loj.ac/problem/6281

思路：

因为是向下取整开方，那么其实每个数最多被开个个几次就会变成0或者1了，更新的时候我们可以将中间的块标记下是否全部变成了0或者1，如果全变了就不处理

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 1e5 + 10;
vector<int>v[M];
int a[M],block,n;
int sum[M];
int tag[M];
int bl[M];
void reset(int x){
    if(tag[x]) return;
    tag[x] = 1;
    sum[x] = 0;
    for(int i = (x-1)*block+1;i <= min(x*block,n);i ++){
        a[i] = sqrt(a[i]);
        sum[x] += a[i];
        if(a[i] > 1) tag[x] = 0;
    }
}

void update(int l,int r){
    for(int i = l;i <= min(bl[l]*block,r);i ++){
        sum[bl[l]] -= a[i];
        a[i] = sqrt(a[i]);
        sum[bl[l]] += a[i];
    }
    if(bl[l] != bl[r]){
        for(int i = (bl[r]-1)*block+1;i <= r;i ++){
            sum[bl[r]] -= a[i];
            a[i] = sqrt(a[i]);
            sum[bl[r]] += a[i];
        }
    }
    for(int i = bl[l] + 1;i <= bl[r] - 1;i ++){
        reset(i);
    }
}

int query(int l,int r){
    int ans = 0;
    for(int i = l;i <= min(bl[l]*block,r);i ++)
        ans += a[i];
    if(bl[l] != bl[r]){
        for(int i = (bl[r]-1)*block+1;i <= r;i ++)
            ans += a[i];
    }
    for(int i = bl[l] + 1;i <= bl[r]-1;i ++)
        ans += sum[i];
    return ans;
}

int main()
{
    int l,r,opt,c;
    scanf("%d",&n);
    block = sqrt(n);
    for(int i = 1;i <= n;i ++){
        scanf("%d",&a[i]);
    }
    for(int i = 1;i <= n;i ++){
        bl[i]  = (i - 1)/block + 1;
        sum[bl[i]] += a[i];
    }
    for(int i = 1;i <= n;i ++){
        scanf("%d%d%d%d",&opt,&l,&r,&c);
        if(opt == 0) update(l,r);
        else printf("%d
",query(l,r));
    }
    return 0;
}