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 2 63.
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
题意：
给你一个含有n个数的数组，m次操作
操作0，给你一个区间l， r 对区间内的每一个数开方。
操作1，输出一个询问区间的数值sum和。
思路：我们看到数据范围是小于等于2^63 ，我们通过本地开方测试可以发现，
数组中的每一个数，最多开方7次，就可以到1 ，， 而1无论咋开方都还是1 ，就是不会改变的数值了。
那么我们用线段树维护区间的sum和，
对于更新，我们暴力更新到线段树的每一个叶子节点，求和就正常的求和。
这里有一点必须的优化就是 如果线段树一个区间的sum和等于区间的长度，这个得出这个区间中的每一个值都是1，那么直接return，不去更新，因为更新是没意义的。
本题2个坑点，：：
1，给的区间x，y ，并不是 x<y 的，有可能 x>y 此处wa多次。
2、题面讲到每一个样例多输出一个回车，刚开始没看到。 此处pe一次。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 100010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n, m;
int root;
ll a[maxn];// 初始点权
ll wt[maxn];// 新建编号点权。
int cnt;// 编号用的变量
int top[maxn];// 所在重链的顶点编号
int id[maxn];//节点的新编号。
std::vector<int> son[maxn];
int SZ[maxn];// 子数大小
int wson[maxn];// 重儿子
int fa[maxn];// 父节点
int dep[maxn];// 节点的深度
struct node
{
int l,r;
ll sum;
ll laze;
}segment_tree[maxn<<2];
void pushup(int rt)
{
segment_tree[rt].sum=(segment_tree[rt<<1].sum+segment_tree[rt<<1|1].sum);
}
void build(int rt,int l,int r)
{
segment_tree[rt].l=l;
segment_tree[rt].r=r;
segment_tree[rt].laze=0;
if(l==r)
{
segment_tree[rt].sum=wt[l];
return;
}
int mid=(l+r)>>1;
build(rt<<1,l,mid);
build(rt<<1|1,mid+1,r);
pushup(rt);
}
void update(int rt,int l,int r)
{
if((segment_tree[rt].l>=l&&segment_tree[rt].r<=r)&&segment_tree[rt].sum==(segment_tree[rt].r-segment_tree[rt].l+1))
{
return ;
}
if(segment_tree[rt].l==segment_tree[rt].r)
{
segment_tree[rt].sum=sqrt(segment_tree[rt].sum);
return ;
}
int mid=(segment_tree[rt].l+segment_tree[rt].r)>>1;
if(mid>=l)
{
update(rt<<1,l,r);
}
if(mid<r)
{
update(rt<<1|1,l,r);
}
pushup(rt);
}
ll query(int rt,int l,int r)
{
if(segment_tree[rt].l>=l&&segment_tree[rt].r<=r)
{
ll res=0ll;
res+=segment_tree[rt].sum;
return res;
}
int mid=(segment_tree[rt].l+segment_tree[rt].r)>>1;
ll res=0ll;
if(mid>=l)
{
res+=query(rt<<1,l,r);
}
if(mid<r)
{
res+=query(rt<<1|1,l,r);
}
return res;
}
int main()
{
// freopen("D:\common_text\code_stream\in.txt","r",stdin);
// freopen("D:\common_text\code_stream\out.txt","w",stdout);
gbtb;
int cas=1;
while(cin>>n)
{
cout<<"Case #"<<cas<<":"<<endl;
repd(i,1,n)
{
cin>>wt[i];
}
build(1,1,n);
cin>>m;
int op;
int x,y;
while(m--)
{
cin>>op>>x>>y;
if(x>y)
{
swap(x,y);
}
if(!op)
{
update(1,x,y);
}else
{
cout<<query(1,x,y)<<endl;
}
}
cout<<endl;
cas++;
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '
');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}