Can you answer these queries? HDU - 4027（线段树）

Can you answer these queries? HDU - 4027

本题是一个看起来像是区间修改的单点修改线段树，因为区间修改不好维护。这个也太难为我了，我想了想好像我不会这种题目，只能暴力求解了，但是问题来了，单点修改的话，绝对超时，但是我不信，写了一发，wa的很开心，然后就想怎么减枝，仔细想了想发现当一个区间的最大值小于等于1的时候就不用修改的，那就很舒服，因为每次是开根号，所以很快就会变成1或0了，时间复杂度是O(能过),值得注意的是x和y的大小是不确定的

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define ms(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x & -x
#define fi first
#define ull unsigned long long
#define se second
#define lson (rt<<1)
#define rson (rt<<1|1)
#define endl "\n"
#define bug cout<<"----acac----"<
#define IOS ios::sync_with_stdio(false), cin.tie(0),cout.tie(0)
using namespace std;
const int maxn = 1e5 + 10;
const int maxm = 1.5e5 + 50;
const double eps = 1e-18;
const double inf = 0x3f3f3f3f;
const ll  lnf = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const  double pi = 3.141592653589;
struct node
{
	ll  l, r, sum,ma;
}a[maxn<<4];
void build(int l,int r,int rt)
{
	a[rt].l = l;
	a[rt].r = r;
	if (l == r)
	{
		scanf("%lld", &a[rt].sum);
		a[rt].ma = a[rt].sum;
		return ;
	}
	int mid = (l + r) >> 1;
	build(l, mid, lson);
	build(mid + 1, r, rson);
	a[rt].sum = a[lson].sum + a[rson].sum;
	a[rt].ma = max(a[lson].ma, a[rson].ma);
}
void update(int L, int R, int rt)
{
	if (a[rt].ma <= 1)return;
	if (a[rt].l ==a[rt] .r)
	{
		a[rt].sum = sqrt(a[rt].sum);
		a[rt].ma = a[rt].sum;
		return;
	}
	int mid = (a[rt].l + a[rt].r) >> 1;
	if (L <= mid)
	{
		update(L, R, lson);
	}
	if (R > mid)
	{
		update(L, R, rson);
	}
	a[rt].sum = a[lson].sum + a[rson].sum;
	a[rt].ma = max(a[lson].ma, a[rson].ma);
}
ll query(int L, int R,int rt)
{
	if (L <= a[rt].l && R >= a[rt].r)
	{
		return a[rt].sum;
	}
	ll ans = 0;
	int mid = (a[rt].l + a[rt].r) >> 1;
	if (L <= mid)
	{
		ans += query(L, R, lson);
	}
	if (R > mid)
	{
		ans += query(L, R, rson);
	}
	return ans;
}
int main()
{
	int cas = 1;
	int n;
	while (~scanf("%d", &n))
	{
		build(1, n, 1);
		int m;
		scanf("%d", &m);
		printf("Case #%d:\n", cas++);
		for (int i = 1; i <= m; i++)
		{
			int T, x, y;
			scanf("%d%d%d", &T, &x, &y);
			if (y < x)swap(x, y);
			if (T == 0)
			{
				update(x, y, 1);
			}
			else
			{
				printf("%lld\n", query(x, y,1));
			}
		}
		printf("\n");
	}
	return 0;
}