HDU 4027 Can you answer these queries? （线段树 区间更新）

题目地址：点击打开链接

题意：给定100000个数，两种操作，0 i j表示将i j这段的数字都开根号（向下取整），1 i j表示查询i j之间的所有值的和（所有的和都不超过64位）

思路：

我们可以很容易的发现，一个数k（k<=2^63-1）在经过最多6，7次的开平方根后，必然会变成1，而且当1的平方根

也是1，也就是说当一个数为1的时候，我们没有必要对它进行操作和更新；而且一个很大的数仅仅经过6，7次就可以

变成1；

所以到这里我们因该就可以形成一个解题的大体思路了：

每当我们要进行更新操作的时候，我们先判断一下这个区间是否有必要进行更新（若全都是1就没有更新的必要了）；

用一个lazy数组记录该区间是否都为1，只有当不都为1的时候才更新，否则不需要更新。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const int maxn = 1e5+5;
ll tree[maxn*4], lazy[maxn*4];//lazy数组记录他的孩子们是否都已经变成了1
int n, m;

void built(int root, int l, int r)
{
    lazy[root] = 0;
    if(l == r)
    {
        scanf("%lld", &tree[root]);
        return;
    }
    int mid = (l+r)/2;
    built(root*2, l, mid);
    built(root*2+1, mid+1, r);
    tree[root] = tree[root*2]+tree[root*2+1];
    lazy[root] = lazy[root*2]&&lazy[root*2+1];
}

ll query(int root, int l, int r, int i, int j)
{
    if(i <= l && j >= r) return tree[root];
    int mid = (l+r)/2;
    ll suml = 0, sumr = 0;
    if(mid >= i) suml = query(root*2, l, mid, i, j);
    if(mid < j) sumr = query(root*2+1, mid+1, r, i, j);
    return suml+sumr;
}

void update(int root, int l, int r, int i, int j)
{
    if(l == r)
    {
        tree[root] = floor(sqrt(tree[root]));
        if(tree[root] <= 1) lazy[root] = 1;
        return ;
    }
    int mid = (l+r)/2;
    if(mid >= i && !lazy[root]) update(root*2, l, mid, i, j);
    if(mid < j && !lazy[root]) update(root*2+1, mid+1, r, i, j);
    tree[root] = tree[root*2]+tree[root*2+1];
    lazy[root] = lazy[root*2]&&lazy[root*2+1];
}

int main(void)
{
    int ca = 1;
    while(cin >> n)
    {
        built(1, 1, n);
        scanf("%d", &m);
        printf("Case #%d:\n", ca++);
        while(m--)
        {
            int cmd, x, y;
            scanf("%d%d%d", &cmd, &x, &y);
            if(x > y) swap(x, y);
            if(cmd) printf("%lld\n", query(1, 1, n, x, y));
            else update(1, 1, n, x, y);
        }
        cout << endl;
    }
    return 0;
}

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 15003    Accepted Submission(s): 3516

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