HDU 4027 Can you answer these queries? （线段树区间求和）

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 12290    Accepted Submission(s): 2912

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

　　简单的线段树区间求和，对数底部的元素开根号的话，这些数不久以后就会变成1，这里就是优化所在了，如果更新时搜到某一区间的和刚好等于这一区间里元素的个数，那么就可以直接返回了（也就是 if(sum[rt] == r - l + 1)  return;）因为对1开根号还是得到了1，不会变了，就不用更新了。时间复杂度也就降下来了。

　　坑点是询问时给你的区间值，前面的可能大于后面，需要swap。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <set>
#include <stack>
#include <cmath>
using namespace std;
#define lson l, m, rt<<1
#define rson m + 1, r, rt<<1|1
#define ll long long
const int maxn = 1e5 + 10;

ll sum[maxn<<2];

void PushUp(int rt) {
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}

void build(int l, int r, int rt) {
    sum[rt] = 0;
    if(l == r) {
        scanf("%lld", &sum[rt]);
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}

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

ll query(int L, int R, int l, int r, int rt) {
    if(L <= l && r <= R) return sum[rt];
    ll res = 0;
    int m = (l + r) >> 1;
    if(L <= m) res += query(L, R, lson);
    if(R > m) res += query(L, R, rson);
    return res;
}

int main() {
    int n;
    int val = 0;
    while(~scanf("%d", &n)) {
        printf("Case #%d:
", ++val);
        build(1, n, 1);
        int m;
        scanf("%d", &m);
        while(m--) {
            int a, b, c;
            scanf("%d %d %d", &a, &b, &c);
            if(c < b) swap(b, c);
            if(a == 0) update(b, c, 1, n, 1);
            else printf("%lld
", query(b, c, 1, n, 1));
        }puts("");
    }
}