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HDU 4027 Can you answer these queries? (线段树 – 区间开根取整)

编程语言 aozil_yang 55℃ 0评论
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题意:

给你n 个数, 两种操作,一种区间成段开根号取整,另一种求区间和。

思路:

因为一个数开上大约7,8次 就成1 了, 1在开根号不变了。

因此我们在递归子区间时,如果发现这个区间都是1了,那就不用在递归了。

在维护一个pushup 求sum 即可

注意:

1. 开long long

2. 区间左右端点可能出现左大于右的情况。

#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
int T;

const int maxn = 100000 + 10;

LL sum[maxn<<2];

void build(int l,int r,int o){
    if (l == r){
        scanf("%I64d",&sum[o]);
        return;
    }
    int m = l + r >> 1;
    build(l,m,o<<1);
    build(m+1,r,o<<1|1);
    sum[o] = sum[o<<1] + sum[o<<1|1];
}

void update(int L,int R,int l,int r,int o){
    if (l == r){
        sum[o] = (LL)sqrt(sum[o]);
        return;
    }
    if (L <= l && r <= R){
        if (sum[o] == (LL)(r-l+1) ){
            return;
        }
    }
    int m = l + r >> 1;
    if (L <= m){
        update(L, R, l, m, o << 1);
    }
    if (m < R) {
        update(L, R, m+1, r, o << 1 | 1);
    }
    sum[o] = sum[o<<1] + sum[o<<1|1];
}

LL query(int L,int R, int l,int r,int o){

    if (L <= l && r <= R){
        return sum[o];
    }
    int m = l + r >> 1;
    LL ans = 0LL;
    if (L <= m){
        ans += query(L, R, l, m, o << 1);
    }
    if (m < R){
        ans += query(L, R, m + 1, r, o << 1 | 1);
    }
    return ans;
}
int ks;
int main(){

    int n;
    while(~scanf("%d",&n)){
        build(1,n,1);
        int q;
        scanf("%d",&q);
        printf("Case #%d:\n", ++ks);
        while(q--){
            int op, x, y;
            scanf("%d %d %d", &op, &x, &y);
            if (x > y) swap(x,y);
            if (op == 0){
                update(x,y,1,n,1);
            }
            else {
                printf("%I64d\n", query(x,y, 1, n, 1));
            }
        }
        putchar('\n');
    }


    return 0;
}

1.Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)


Total Submission(s): 16232    Accepted Submission(s): 3802







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
 





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