L - Can you answer these queries?

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 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：对一串数字某一区间进行开根操作或者求某一区间和

解法：这里有一个非常重要的信息，即和最大为2的63次方，而开七次方就可以一定将他开为1，所以我们可以采用单点更新，设置一个标记，判断该区间是否为0或1，是的话就不用更新了，复杂度为nlogn

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;
#define MAXN 100010
long long a[MAXN];
struct node
{
    int l,r,sign;//sign用来标记
    long long sum;
}tr[MAXN*4];
void build(int id,int l,int r)
{
    tr[id].l=l;
    tr[id].r=r;
    tr[id].sign=1;//初始化
    if(l==r)
    {
        tr[id].sum=a[l];
        if(a[l]==0||a[l]==1)
            tr[id].sign=0;//为零或一则变为零
    }
    else
    {
        int mid=(l+r)/2;
        build(id*2,l,mid);
        build(id*2+1,mid+1,r);
        tr[id].sign=tr[id*2].sign|tr[id*2+1].sign;//父亲节点标记由儿子节点递推出来
        tr[id].sum=tr[id*2].sum+tr[id*2+1].sum;
    }
}
void update(int id,int l,int r)//单点更新
{
    if(tr[id].sign==0)//表明该区间全为小于2的数，直接返回
        return ;
    if(tr[id].l==tr[id].r)
    {
        tr[id].sum=(long long)sqrt(tr[id].sum);
        if(tr[id].sum==1)
            tr[id].sign=0;
        return;
    }
    else
    {
        int mid=(tr[id].l+tr[id].r)/2;
        if(r<=mid)
            update(id*2,l,r);
        else if(l>mid)
            update(id*2+1,l,r);
        else
        {
            update(id*2,l,mid);
            update(id*2+1,mid+1,r);
        }
        tr[id].sign=tr[id*2].sign|tr[id*2+1].sign;//每次都需要更新
        tr[id].sum=tr[id*2].sum+tr[id*2+1].sum;
    }
}
long long query(int id,int l,int r)
{
    if(tr[id].l>=l&&tr[id].r<=r)
        return tr[id].sum;
    else
    {
        int mid=(tr[id].l+tr[id].r)/2;
        if(r<=mid)
            return query(id*2,l,r);
        else if(l>mid)
            return query(id*2+1,l,r);
        else
            return query(id*2,l,mid)+query(id*2+1,mid+1,r);
    }
}
int main()
{
    int n,m,i,x,y,z,t=0;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        build(1,1,n);
        scanf("%d",&m);
        t++;
        printf("Case #%d:\n",t);
        while(m--)
        {
            scanf("%d%d%d",&x,&y,&z);
            if(x==0)
            {
                if(y>z)
                {
                    int w;
                    w=y;
                    y=z;
                    z=w;
                }
                update(1,y,z);
            }
            else
            {
                if(y>z)
                {
                    int w;
                    w=y;
                    y=z;
                    z=w;
                }
                printf("%lld\n",query(1,y,z));
            }
        }
        printf("\n");
    }
    return 0;
}