# Sprinkler (1), sprinkler (

Source: Internet
Author: User

Sprinkler (1), sprinkler (

Sprinkler (1) Time Limit: 3000 MS | memory limit: 65535 KB difficulty: 3
Description
There is a lawn with a length of 20 meters and a width of 2 meters. A sprinkler with a radius of Ri should be placed on the horizontal center, the effect of each sprinkler will make the circle with its center radius as the real number Ri (0 <Ri <15) moist, there are plenty of sprinkler I (1 <I <600), and you will surely be able to wet all the lawns. What you need to do is: select as few sprinkler as possible, wet all the lawns.
Input
The first line m indicates that there are m groups of test data
The first line of each set of test data has an integer n, n indicates a total of n sprinkler devices, and the subsequent line has n real number ri, ri indicates the radius of the circle covered by the sprinkler.
Output
Number of devices used for output
Sample Input
`252 3.2 4 4.5 6 101 2 3 1 2 1.2 3 1.1 1 2`
Sample output
`25`

This problem simplifies many operations because the width of the matrix is 2...

You only need to calculate the oblique edge length of the lawn, and then as long as the radius of all circles is larger than half of the length of the Oblique Edge (but you must discard the device with a radius less than or equal to 1, no matter how it is placed on the horizontal line, it cannot completely cover the grass ).

X = sqrt (r * r-(h/2) * (h/2) so that the circle with a radius of r can overwrite the length of 2x, with the remaining w-2x, in this way, when w <= 0, it is satisfied.

R = sqrt (x * x + h * h)/2 Total r = sqrt (w * w + h * h)/2

Simply put, the sum of the radius (not equal to or less than 1) of all circles is greater than half of the length of the Oblique Edge.

`#include <iostream>02.#include <algorithm>03.#include <cmath>04.#include <cstdio>05.using namespace std;06.const double PI = acos(-1.0);07.int main()08.{09.int n,m;10.cin>>n;11.while(n--)12.{13.double a[610];14.int m;15.cin>>m;16.for(int i = 0; i < m; i++) cin>>a[i];17.sort(a,a+m);18.double sum = 0;19.double len = sqrt(20*20 + 2*2)/2;20.int count = 0;21.for(int i = m-1; i >=0 ;i--)22.{23.sum += a[i];24.count++;25.if(sum > len) break;26.}27.cout<<count<<endl;28.}29.return 0;30.}`

Related Keywords:

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

## A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• #### Sales Support

1 on 1 presale consultation

• #### After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.