HDU4027(线段树单点更新区间)

Can you answer these queries?

Crawling in process... Crawling failed Time Limit:2000MS     Memory Limit:65768KB     64bit IO Format:%I64d & %I64uTime Limit:2000MS     Memory Limit:65768KB     64bit IO Format:%I64d & %I64u

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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

优化:当子树的节点值为1时无需再更新;

 

 

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN=100005;
typedef long long ll;
struct Node{
    ll sum;
    int l,r;
}a[MAXN*4];
void build(int rt,int l,int r)
{
    a[rt].l=l;
    a[rt].r=r;
    if(l==r)
    {
        scanf("%lld",&a[rt].sum);
        return ;
    }
    int mid=(l+r)>>1;
    build(rt<<1,l,mid);
    build((rt<<1)|1,mid+1,r);
    a[rt].sum=a[rt<<1].sum+a[(rt<<1)|1].sum;
}
void update(int rt,int l,int r)
{
    if(a[rt].r-a[rt].l+1==a[rt].sum)    return ;// 优化
    if(a[rt].l==a[rt].r)
    {
        a[rt].sum=sqrt(1.0*a[rt].sum);
        return ;
    }
    int mid=(a[rt].l+a[rt].r)>>1;
    if(r<=mid)
        update(rt<<1,l,r);
    else if(mid<l)
        update((rt<<1)|1,l,r);
    else
    {
        update(rt<<1,l,mid);
        update((rt<<1)|1,mid+1,r);
    }
    a[rt].sum=a[rt<<1].sum+a[(rt<<1)|1].sum;
}
ll query(int rt,int l,int r)
{
    if(a[rt].l==l&&a[rt].r==r)
    {
        return a[rt].sum;
    }
    int mid=(a[rt].l+a[rt].r)>>1;
    if(r<=mid)
        return query(rt<<1,l,r);
    else if(mid<l)
        return query((rt<<1)|1,l,r);
    else
    {
        return query(rt<<1,l,mid)+query((rt<<1)|1,mid+1,r);
    }
}
int n,m;
int main()
{
    int cas=1;
    while(scanf("%d",&n)!=EOF)
    {
        build(1,1,n);
        printf("Case #%d:
",cas++);
        scanf("%d",&m);
        while(m--)
        {
            int op,x,y;
            scanf("%d%d%d",&op,&x,&y);
            if(x>y)    swap(x,y);//x可能比y大
            if(op==0)
                update(1,x,y);
            else
                printf("%lld
",query(1,x,y));
        }
        printf("
");
    }
    
    return 0;
}