hdu 4027 Can you answer these queries?

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define N 100010
#define Lson root<<1, L, tree[root].Mid()
#define Rson root<<1|1, tree[root].Mid() + 1, R

using namespace std;

struct Tree
{
    int L, R;
    long long sum;
    int Mid()
    {
        return (L + R) / 2;
    }
    int Len()
    {
        return (R - L + 1);
    }
} tree[N * 4];

long long al[N];

void Build(int root, int L, int R)
{
    tree[root].L = L, tree[root].R = R;
    if(L == R)
    {
        tree[root].sum = al[L];
        return ;
    }

    Build(Lson);
    Build(Rson);

    tree[root].sum = tree[root<<1].sum + tree[root<<1|1].sum;
}

void Update(int root, int L, int R)
{
    if(tree[root].Len() == tree[root].sum)
        return ;//注意这儿如果省了会TLE
    if(tree[root].L == tree[root].R)
    {
        tree[root].sum = (long long)sqrt(tree[root].sum);
        return ;
    }
    if(R <= tree[root].Mid())
        Update(root<<1, L, R);
    else if(L > tree[root].Mid())
        Update(root<<1|1, L, R);
    else
    {
        Update(Lson);
        Update(Rson);
    }
    tree[root].sum = tree[root<<1].sum + tree[root<<1|1].sum;
}

long long Query(int root, int L, int R)
{
    if(tree[root].L == L && tree[root].R == R)
        return tree[root].sum;
    if(R <= tree[root].Mid())
        return Query(root<<1, L, R);
    else if(L > tree[root].Mid())
        return Query(root<<1|1, L, R);
    else
        return Query(Lson) + Query(Rson);
}

int main()
{
    int n, m, i, t, a, b, x = 0;
    while(~scanf("%d", &n))
    {
        x++;
        for(i = 1 ; i <= n ; i++)
            scanf("%I64d", &al[i]);
        Build(1, 1, n);
        scanf("%d", &m);
        printf("Case #%d:\n", x);
        while(m--)
        {
            scanf("%d%d%d", &t, &a, &b);
            if(a > b)
                swap(a, b);//注意，如果所给的a比b大，两者就需要调换位置
            if(t == 0)
                Update(1, a, b);
            else
                printf("%I64d\n", Query(1, a, b));
        }
        printf("\n");
    }
    return 0;
}