Sprinkler (1)

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 sample input

252 3.2 4 4.5 6 101 2 1 2 1.2 3 1.1 1 2

Sample output

2
5

Understanding of the details: N indicates a total of N sprinkler devices. This sentence means that there is a corresponding device for each radius R, there are n devices, of course, corresponding to n r; at first it was understood that we can always wet the lawn with the largest radius, and we can find that each device can only be used once. the following code and comments of the greedy method are provided;

#include 
      
        # include 
       
         // contains the rapidly sorted header file # include 
        
          // mathematical function header file double length (double r) // convert R to a valid value {return 2 * SQRT (double) (R * r-1 ));} int COM (const void * a, const void * B) // The Quick Sort comparison function {return (* (double *) B-* (double *) a> 0? 1:-1); // sort from big to small} int main () {int m, n, I, j = 0; // J counts the number of devices that meet> 1. Double R [1020], temp, sum = 0.0; // stores the scanf ("% d ", & M); While (M --) {scanf ("% d", & N); for (I = 0, j = 0; I 
         
           1) // discard devices less than or equal to 1 in radius R [J ++] = length (temp);} qsort (R, J, sizeof (R [0]), com ); // for (I = 0; I 
          
            = 20) // If sum is greater than or equal to 20, it indicates that break has been wet ;} printf ("% d \ n", I + 1); // Add 1 to the output, corresponding to the relationship with I, if I = 0 is to exit the loop, the task can be completed by one device. The relationship between the number of devices and I is displayed.} return 0;} 
          
         
        
       
      

sprinkler (1)