Nyoj 6 sprinkler (1)

/*
Question:
Chinese Translation:
Question:
Solution:
Difficulties: Use a formula SQRT (A [I] * A [I]-1 ). He asks for the side length, so that the side length is sum, sum = SQRT (R * R-1) * 2, if not enough to continue adding a rectangle know enough.

Key points: fast release,
Problem solving person: lingnichong
Solution time:

Feeling: I have been writing it for a long time before I can understand what it means.
*/

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

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>//#include<algorithm>//using namespace std;double a[660]={0};int cmp(const void *b,const void *a){return *(double *)a>*(double *)b? 1:-1;}int main(){int n,m,i,j;double sum;scanf("%d",&n);while(n--){memset(a,0,sizeof(a));scanf("%d",&m);for(i=0;i<m;i++)scanf("%lf",&a[i]);//sort(a,a+m);qsort(a,m,sizeof(a[0]),cmp);sum=0.0;for(i=0;i<m&&sum<10;i++)sum+=sqrt(a[i]*a[i]-1);printf("%d\n",i);}return 0;}