分块（区间根号，区间查询

题目大意就是给你一个区间在区间的各个数都开个根号，然后在让你区间求和

题目链接https://loj.ac/problem/6281

主要思路就是如果需更新区间全是0或者全是1就可以不用管。这题我用了set，头尾指针指的值相等且为0或者1就满足条件。

#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 color[maxn], l[maxn], r[maxn], pos[maxn], v[maxn], sum[maxn], lz[maxn], t;
set<int> s[1010];

inline int read() //快读
{
    char ch = getchar(); ll k = 0, f = 1;
    while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9') {k = (k << 1) + (k << 3) + ch - '0'; ch = getchar();}
    return k * f;
}

inline void reset(int p)
{
    s[p].clear();
    s[p].insert(v + l[p], v + r[p] + 1);  //自动从小到大排序
    set<int>::iterator it, itt;
    it = s[p].begin(); itt = s[p].end(); itt--;   //左闭右开，否则s.end()返回的应该是s.size()
    if(sqrt(*it) == 1 && sqrt(*itt) == 1) color[p] = 1;
    if(sqrt(*it) == 0 && sqrt(*itt) == 0) color[p] = 1;  
}

inline void inser(int a, int b)
{
    int pa = pos[a], pb = pos[b];
    if(pa == pb)
    {
        for(int i = a; i <= b; ++i)
        {
            sum[pa] -= v[i];
            v[i] = sqrt(v[i]);
            sum[pa] += v[i];
        }
        reset(pa);
    }
    else
    {
        for(int i = a; i <= r[pa]; ++i)
        {
            sum[pa] -= v[i];
            v[i] = sqrt(v[i]);
            sum[pa] += v[i];
        }
        reset(pa);

        for(int i = l[pb]; i <= b; ++i)
        {
            sum[pb] -= v[i];
            v[i] = sqrt(v[i]);
            sum[pb] += v[i];
        }
        reset(pb);

        for(int i = pa + 1; i <= pb - 1; ++i)
        {
            if(color[i]==0)  //如果不是全为1或者0的区间就要进行更新
            {
                for(int j = l[i]; j <= r[i]; ++j)
                {
                    sum[i] -= v[j];
                    v[j] = sqrt(v[j]);
                    sum[i] += v[j];
                }
                reset(i);
            }
        }

    }
}

inline int findy(int a, int b)
{
    int pa = pos[a], pb = pos[b], ans = 0;
    if(pa == pb)
    {
        for(int i = a; i <= b; ++i) ans += v[i];
    }
    else
    {
        for(int i = a; i <= r[pa]; ++i) ans += v[i];
        for(int i = l[pb]; i <= b; ++i) ans += v[i];
        for(int i = pa + 1; i <= pb - 1; ++i)
        {
            ans += sum[i];
        }
    }
    return ans;
}

int main()
{
    int n, opt, a, b, c;
    n = read();
    for(int i = 1; i <= n; ++i) v[i] = read();
    t = sqrt(n);
    for(int i = 1; i <= t; ++i)
    {
        l[i] = (i-1) * t + 1;
        r[i] = i * t;
    }

    if(r[t] != n) t++, l[t] = r[t - 1] + 1, r[t] = n;

    for(int i = 1; i <= t; ++i)
    {
        for(int j = l[i]; j <= r[i]; ++j)
        {
            pos[j] = i; s[i].insert(j); sum[i] += v[j];  //初始化
        }
    }

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

    return 0;
}

另外需要提下的是RTE和TLE不一定是正常情况，还有可能是用了快读又关了同步流，即getchar()和scanf printf类似，用cin cout再用这些都不能关闭同步流。（直接用scanf printf就没这么多事

虽然没发生在我身上。。但是一直检查对的代码也是很痛苦的

也告诉我们学东西要学透彻别老想一知半解