HDU 4027 Can you answer these queries （有意思的线段树）

题意: 区间开方, 然后求区间的和

分析: 乍一看像是区间开方 无法维护这个和 但是由于题目说了 维护的是整数(round down to integer) 

          我们可以想到 开方的次数不会很多 这样我们直接暴力单点更新 也没有问题啦 问题解决

代码:

//
//  Created by TaoSama on 2015-09-16
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, q;
long long sum[N << 2];

#define root 1, n, 1
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

void push_up(int rt) {
sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void build(int l, int r, int rt) {
if(l == r) {
scanf("%I64d", &sum[rt]);
return;
}
int m = l + r >> 1;
build(lson);
build(rson);
push_up(rt);
}

void update(int L, int R, int l, int r, int rt) {
if(sum[rt] == r - l + 1) return;
if(l == r) {
sum[rt] = sqrt(sum[rt]);
return;
}
int m = l + r >> 1;
if(L <= m) update(L, R, lson);
if(R > m) update(L, R, rson);
push_up(rt);
}

long long query(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
return sum[rt];
}
int m = l + r >> 1;
long long ret = 0;
if(L <= m) ret += query(L, R, lson);
if(R > m) ret += query(L, R, rson);
return ret;
}

int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int kase = 0;
while(scanf("%d", &n) == 1) {
build(root);
scanf("%d", &q);
printf("Case #%d:\n", ++kase);
while(q--) {
int op, l, r; scanf("%d%d%d", &op, &l, &r);
if(l > r) swap(l, r);
if(op) printf("%I64d\n", query(l, r, root));
else update(l, r, root);
}
puts("");
}
return 0;
}