Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 8674 Accepted Submission(s): 1971


Problem Description A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help.
You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line.

Notice that the square root operation should be rounded down to integer.
Input The input contains several test cases, terminated by EOF.
For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000)
The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 2⁶³.
The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)
For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive.

Output For each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.
Sample Input 10 1 2 3 4 5 6 7 8 9 10 5 0 1 10 1 1 10 1 1 5 0 5 8 1 4 8
Sample Output Case #1: 19 7 6
Source 0：區間每個數開平方根取整。 1：區間和

#include<stdio.h>
#include<math.h>
#define N 100100
struct nn
{
    int flag;//為0，說明這個區間內的數全為1
    __int64 sum;//區間和
}tree[N*3];
void builde(int l,int r,int k)
{
    int m=(l+r)/2;
    if(l==r)
    {
        scanf("%I64d",&tree[k].sum);
      tree[k].flag=1;if(tree[k].sum<=1)tree[k].flag=0; return ;
    }
    builde(l,m,k*2);
    builde(m+1,r,k*2+1);
    tree[k].sum=tree[k*2].sum+tree[k*2+1].sum;
    tree[k].flag=tree[k*2].flag|tree[k*2+1].flag;
}
void updata(int l,int r,int k,int L,int R)
{
    int m=(r+l)/2;
    if(l==r)
    {
        tree[k].sum=(__int64)sqrt(1.0*tree[k].sum);
        if(tree[k].sum<=1)tree[k].flag=0; return ;
    }
    if(L<=m&&tree[k*2].flag)updata(l,m,k*2,L,R);
    if(R>m&&tree[k*2+1].flag)updata(m+1,r,k*2+1,L,R);
    tree[k].sum=tree[k*2].sum+tree[k*2+1].sum;
    tree[k].flag=tree[k*2].flag|tree[k*2+1].flag;
}
__int64 query(int l,int r,int k,int L,int R)
{
    int m=(l+r)/2;
    if(L<=l&&r<=R)
    {
        return tree[k].sum;
    }
    __int64 sum=0;
    if(L<=m)sum+=query(l,m,k*2,L,R);
    if(R>m)sum+=query(m+1,r,k*2+1,L,R);
    return sum;
}
int main()
{
    int n,m,t=0,o,L,R;
    while(scanf("%d",&n)>0)
    {
        builde(1,n,1);
        scanf("%d",&m);
        printf("Case #%d:\n",++t);
        while(m--)
        {
            scanf("%d%d%d",&o,&L,&R);
            int tem;
            if(L>R){tem=L;L=R;R=tem;}//注意了
            if(o==0)updata(1,n,1,L,R);
            else
                printf("%I64d\n",query(1,n,1,L,R));
        }
        printf("\n");
    }
    return 0;
}