Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 13517    Accepted Submission(s): 3081


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 //真TMD無語 //又寫一道 單點更新的線段樹 //本來 以為 Soeasy TMD 寫出來又 TEL // 用延遲標記了啊 // 原來這道題 就是 開根號到 1 就不需要在 update下去了

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
using namespace std;

const int L = 100000+10;
__int64 sum[L<<2],cnt[L<<2];

void add(int i)
{
    sum[i] = sum[2*i]+sum[2*i+1];
    cnt[i] = cnt[2*i]&&cnt[2*i+1];
}

void init(int l,int r,int i)
{
    if(l == r)
    {
        scanf("%I64d",&sum[i]);
        return;
    }
    int mid = (l+r)>>1;
    init(l,mid,2*i);
    init(mid+1,r,2*i+1);
    add(i);
}

void insert(int L,int R,int l,int r,int i)
{
    if(l == r)
    {
        sum[i] = sqrt(sum[i]);//向下取整
        if(sum[i]<=1)//如果為0，則標記不能再訪問了
            cnt[i] = 1;
        return ;
    }
    int mid = (l+r)>>1;
    if(L<=mid && !cnt[2*i])//孩子已經破壞了則不需再訪問
        insert(L,R,l,mid,2*i);
    if(R>mid && !cnt[2*i+1])
        insert(L,R,mid+1,r,2*i+1);
    add(i);
}

__int64 query(int L,int R,int l,int r,int i)//查詢區間和
{
    if(L<=l && r<=R)
        return sum[i];
    int mid = (l+r)>>1;
    __int64 ans = 0;
    if(L<=mid)
        ans+=query(L,R,l,mid,2*i);
    if(R>mid)
        ans+=query(L,R,mid+1,r,2*i+1);
    return ans;
}

int main()
{
    int n,m,i,j,x,l,r,cas = 1,flag;
    while(~scanf("%d",&n))
    {
        memset(sum,0,sizeof(sum));
        memset(cnt,0,sizeof(cnt));
        init(1,n,1);
        printf("Case #%d:\n",cas++);
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d%d%d",&flag,&l,&r);
            if(l>r)//這裡要注意
                swap(l,r);
            if(flag)
                printf("%I64d\n",query(l,r,1,n,1));
            else
                insert(l,r,1,n,1);
        }
        printf("\n");
    }

    return 0;
}