[pat (basic level) practice] - [simple simulation] 1063 calculate the spectral radius

One 【 questions 】

• Class B

Two 【 Title number 】

• 1063 Calculate the spectral radius (20 branch )

3、 ... and 【 Title Description 】

• In mathematics , Matrix “ Spectral radius ” Is the supremum of the modular set of its eigenvalues . In other words , For a given $n$ Eigenvalues of a complex space $a_1+b_{1}i,\cdots ,a_n+b_{n}i$, Their modules are the square of the sum of squares of real and imaginary parts , and “ Spectral radius ” Is the maximum module .
• Now let's give some eigenvalues of complex space , Please calculate and output the spectral radius of these eigenvalues .

Four 【 Title Example 】

• Input format ：
  Enter the first line to give a positive integer N（≤ 10 000） Is the number of input eigenvalues . And then N That's ok , Each line gives 1 The real and imaginary parts of eigenvalues , Separated by spaces . Be careful ： The problem is to ensure that both the real part and the imaginary part are absolute values that do not exceed 1000 The integer of .

• Output format ：
  Output the spectral radius in one line , Round to decimal 2 position .

• sample input ：
  5
  0 1
  2 0
  -1 0
  3 3
  0 -3

• sample output ：
  4.24

5、 ... and 【 Their thinking 】

• The first thing that came to my mind , According to the requirements of the topic, the analog output is then calculated and stored in the array , Then sort the array from large to small , The first is the output , But there are always two use cases that fail , You have to 14 branch , I didn't think about the specific reason . Later, it was found that there was no need for storage , Because we just need to find the biggest , So first set up $max=0$, Because the smallest is $0$, If the result of analog operation is better than $max$ Big update , After traversing the output data, you can get the final result

6、 ... and 【 The final score 】

• 20 branch

7、 ... and 【 Code implementation 】

#include<stdio.h>
#include<math.h>
int main()
{
    
    int n;
    scanf("%d",&n);
    double max = 0;
    for(int i = 0;i<n;i++)
    {
    
        double n1,n2;
        scanf("%lf %lf",&n1,&n2);
        double temp = sqrt(pow(n1,2) + pow(n2,2));
        if(temp > max)
        {
    
            max = temp;
        }
    }
    printf("%.2lf",max);
    return 0;
}

8、 ... and 【 Submit results 】