Can you answer these queries (HDU 4027)

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https://vjudge.net/problem/HDU-4027

题意：给一个数组序列，数组长度为100000，现在有两种操作，
一种操作是将某一个固定区间所有数开方（向下取整），
另一种操作是询问某个区间的所有数字之和。

思路：当sum[rt] = r-l+1的时候就不需要往下递归了，其他时候照常

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <cmath>#include <stack>#include <string>#include <sstream>#include <map>#include <set>#define pi acos(-1.0)#define LL long long#define ULL unsigned long long#define inf 0x3f3f3f3f#define INF 1e18#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;typedef pair<int, int> P;const double eps = 1e-10;const int maxn = 1e5 + 5;const int N = 1e4 + 5;const int mod = 1e8;LL sum[maxn<<2]; void Build(int l, int r, int rt){     if (l == r){        scanf("%I64d", &sum[rt]);        return;    }    int mid = (l + r) >> 1;    Build(lson);    Build(rson);    sum[rt] = sum[rt<<1] + sum[rt<<1|1];} LL Query(int L, int R, int l, int r, int rt){    if (L<=l && r<=R)        return sum[rt];     int mid = (l + r) >> 1;    LL res = 0;    if (L <= mid) res += Query(L, R, lson);    if (R > mid)  res += Query(L, R, rson);    return res;}void Update(int L, int R, int l, int r, int rt){    if (L<=l && r<=R && sum[rt] == r-l+1){        return;    }    if (l == r){        sum[rt] = sqrt(sum[rt]);        return ;    }     int mid = (l + r) >> 1;    if (L <= mid) Update(L, R, lson);    if (R > mid)  Update(L, R, rson);    sum[rt] = sum[rt<<1] + sum[rt<<1|1];}int main(void){//    freopen("in.txt", "r", stdin);    int n, m, cas = 1;    while (cin >> n){        Build(1, n, 1);        cin >> m;        printf("Case #%d:\n", cas++);        for (int i = 1; i <= m; i++){            int T, l, r;            cin >> T >> l >> r;            int a = min(l, r);            int b = max(l, r);            if (T == 0){                Update(a, b, 1, n, 1);            }            else {                cout << Query(a, b, 1, n, 1) << endl;            }        }        cout << endl;    }return 0;}

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