Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 19391 Accepted Submission(s): 4584

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
2 3 4 5 6 7 8 9 10
5
1 10
1 10
1 5
5 8
4 8

Sample Output

Case #1:
19
7
6

/*
題意：
給你n個數，n<=100000，每個數不超過2^63
兩種操作
0 x y 表示從x到y每個數都開方取下正
1 x y 表示從x到y每個數求和

題解：
2^63開方 6~7次變爲1之後就不用操作了,也就是當sum==區間長度時候
這道題類似hdu5239平方後mod 9223372034707292160 即2 ^ 63 - 2 ^ 31
即任意數的平方MOD此數，重複操作至多29次就會進入一個不變的數。

*/

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <cmath>
using namespace std;
typedef long long ll;

const int maxn = 101000*4;
const int INF = 0x3f3f3f3f;

struct node{
    int l,r;
    ll sum;
}tree[maxn];

void build(int node,int b,int e){
    int mid = (b+e)/2;
    tree[node].l = b;
    tree[node].r = e;

    if(b==e) {
       scanf("%lld",&tree[node].sum);
       return;
    }
    build(node*2,b,mid);
    build(node*2+1,mid+1,e);
    tree[node].sum = tree[node*2].sum + tree[node*2+1].sum;
}


void update(int node ,int ql,int qr){
    if(tree[node].sum == (tree[node].r-tree[node].l+1)) return;
    if(tree[node].l == tree[node].r){
        tree[node].sum = (ll)sqrt(tree[node].sum);
        return;
    }

    int mid  = (tree[node].l + tree[node].r)/2;
    if(ql<=mid) update(node*2,ql,qr);
    if(qr>mid) update(node*2+1,ql,qr);
    tree[node].sum = tree[node*2].sum + tree[node*2+1].sum;
}

ll query(int node,int ql,int qr){
    ll ans1 = 0,ans2 = 0;
    if(ql<=tree[node].l && qr>=tree[node].r)
        return tree[node].sum;

    int mid = (tree[node].l + tree[node].r)/2;
    if(ql<=mid) ans1 = query(node*2,ql,qr);
    if(qr>mid) ans2 = query(node*2+1,ql,qr);
    return ans1+ans2;
}


int main(){
    int n,m;
    int icase = 1;
    while(scanf("%d",&n)!=EOF){
        printf("Case #%d:\n",icase++);
        build(1,1,n);
        scanf("%d",&m);
        while(m--){
            int t,x,y;
            scanf("%d%d%d",&t,&x,&y);
            if(x>y) swap(x,y);///這裏有坑
            if(t==0) update(1,x,y);
            if(t==1) printf("%lld\n",query(1,x,y));
        }
        puts("");
    }

    return 0;
}