HDU 4027 - Can you answer these queries?(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): 15043 Accepted Submission(s): 3529

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

題意:
給出一個數列, 有更新和區間求和的操作.

更新要求, 區間內的所有數都要開平方.

解題思路:
1.區間更新, 用lazy標記, 如果當前點被更新次數大於5, 那麼結果就爲1(因爲題目給的是1-2^63).

2.單點更新, 如果當前的數已經是1, 那麼就沒有必要更新了.
即當前區間的區間和爲r - l + 1, 那麼區間內的所有點都爲1, 就不用更新.

AC代碼:

#include <stdio.h>
#include <algorithm>
#include <cmath>
#define int long long
#define lson l, mid, rt << 1
#define rson mid+1, r, rt << 1 | 1
using namespace std;
const int maxn = 1e5+5;
int segTree[maxn << 2];

void pushUp(int rt)
{
    segTree[rt] = segTree[rt << 1] + segTree[rt << 1 | 1];
}

void build(int l, int r , int rt)
{
    if(l == r){
        scanf("%lld",&segTree[rt]);
        return ;
    }
    int mid = (l+r) >> 1;
    build(lson);
    build(rson);
    pushUp(rt);
}

void update(int L, int R, int l, int r, int rt)
{
    if(L <= l && R >= r && segTree[rt] == r - l + 1)
        return ;
    if(l == r){
        segTree[rt] = sqrt((double) segTree[rt]);
        return ;
    }
    int mid = (l+r) >> 1;
    if(L <= mid)
        update(L, R, lson);
    if(R > mid)
        update(L, R, rson);
    pushUp(rt);
}

int query(int L, int R, int l, int r, int rt)
{
    if(L <= l && R >= r)
        return segTree[rt];
    int mid = (l+r) >> 1;
    int res = 0;
    if(L <= mid)
        res += query(L, R, lson);
    if(R > mid)
        res += query(L, R, rson);
    return res;
}
void main()
{
    int n;
    int c = 1;
    while(~scanf("%lld",&n)){
        build(1, n, 1);
        int t;
        scanf("%lld",&t);
        printf("Case #%lld:\n",c++);
        while(t--){
            int op;
            int a;
            int b;
            scanf("%lld%lld%lld",&op, &a, &b);
            if(a > b)
                swap(a, b);
            if(op)
                printf("%lld\n",query(a, b, 1, n, 1));
            else
                update(a, b, 1, n, 1);
        }
        putchar('\n');
    }
}