HDU4027 线段树

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 15314    Accepted Submission(s): 3590

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

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

 题意：

n个数，两种操作：0 b c表示将[b,c]区间的数开方，1 b c表示询问[b,c]区间的和。

代码：

//因为题目中所有和不超过2^63，所以每个数开方次数不超过7，这样在更新[b,c]时
//如果区间内的所有数都是1就不必更新了，否则挨个更新叶子。
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
const int maxn=100005;
ll sum[maxn*4+10];
void Pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void Build(int l,int r,int rt)
{
    if(l==r){
        scanf("%lld",&sum[rt]);
        return;
    }
    int m=(l+r)>>1;
    Build(l,m,rt<<1);
    Build(m+1,r,rt<<1|1);
    Pushup(rt);
}
void Update(int ql,int qr,int l,int r,int rt)
{
    if(l==r){
        sum[rt]=sqrt(sum[rt]);
        return;
    }
    if(ql<=l&&qr>=r&&sum[rt]==r-l+1) return;
    int m=(l+r)>>1;
    if(ql<=m) Update(ql,qr,l,m,rt<<1);
    if(qr>m) Update(ql,qr,m+1,r,rt<<1|1);
    Pushup(rt);
}
ll Query(int ql,int qr,int l,int r,int rt)
{
    if(ql<=l&&qr>=r) return sum[rt];
    int m=(l+r)>>1;
    ll s=0;
    if(ql<=m) s+=Query(ql,qr,l,m,rt<<1);
    if(qr>m) s+=Query(ql,qr,m+1,r,rt<<1|1);
    return s;
}
int main()
{
    int n,m,cas=0;
    while(scanf("%d",&n)==1){
        Build(1,n,1);
        scanf("%d",&m);
        int a,b,c;
        printf("Case #%d:
",++cas);
        while(m--){
            scanf("%d%d%d",&a,&b,&c);
            int bb=min(b,c),cc=max(b,c);
            if(a) printf("%lld
",Query(bb,cc,1,n,1));
            else Update(bb,cc,1,n,1);
        }
        printf("
");
    }
    return 0;
}