HDU4027 線段樹

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 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

 

 

Source

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

 題意：

n個數，兩種操作：0 b c表示將[b,c]區間的數開方，1 b c表示詢問[b,c]區間的和。

代碼：

//因為題目中所有和不超過2^63，所以每個數開方次數不超過7，這樣在更新[b,c]時
//如果區間內的所有數都是1就不必更新了，否則挨個更新葉子。
#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:\n",++cas);
        while(m--){
            scanf("%d%d%d",&a,&b,&c);
            int bb=min(b,c),cc=max(b,c);
            if(a) printf("%lld\n",Query(bb,cc,1,n,1));
            else Update(bb,cc,1,n,1);
        }
        printf("\n");
    }
    return 0;
}