# Average performance

Problem Description
Suppose that there are n (n < = 50) students in a class, and each of them takes m (m < = 5) courses. The average score of each student and the average score of each course are calculated, and the number of students whose scores of each subject are greater than or equal to the average is output.

Input
The input data has multiple test instances. The first line of each test instance includes two integers n and m, representing the number of students and courses respectively. Then there are n rows of data, each row including M integers (i.e. test scores).

Output
For each test instance, three rows of data are output, the first row contains n data, which represents the average score of n students, and the result retains two decimal places; the second row contains m data, which represents the average score of m courses, and the result retains two decimal places; the third row is an integer, which represents the number of students whose scores of each subject in the class are greater than or equal to the average score.
Each test instance is followed by a blank line.

Sample Input
2 2
5 10
10 20

Sample Output
7.50 15.00
7.50 15.00
1

``` 1 #include<iostream>
2 #include<cstdio>
3 using namespace std;
4
5 int main()
6 {
7     double m,n;
8     while(cin >> n >> m)//n A student, m Subjects
9     {
10         double fenshu = {0},fen = {0};//fenshu Two dimensional array records the scores of each student, fen One dimensional array records the total score of a student
11         for(int i = 0;i < n;i++)
12         {
13             for(int j = 0;j < m;j++)
14             {
15                 cin >> fenshu[i][j];//Cyclic read in n*m Score data
16                 fen[i] += fenshu[i][j];//The scores of each student are accumulated and recorded in one-dimensional array fen in
17             }
18         }
19         //Output first line
20         for(int i = 0;i < n;i++) //Pay attention to control the output format, and calculate the average score of each student
21         {
22             if(i == n - 1)
23                 printf("%.2lf",fen[i] / m);
24             else
25                 printf("%.2lf ",fen[i] / m);
26         }
27         cout << endl;
28         //Output second line
29         double ave1 = {0},ave2 = {0};//ave1 Array records the total score of each department, ave2 Array record the average score of each subject, pay attention to the format of output
30         for(int j = 0;j < m;j++)//The outer cycle is j Equivalent to summing up a column
31         {
32             for(int i = 0;i < n;i++)//Inner circulation is i//First, the scores of each subject are accumulated
33                 ave1[j] += fenshu[i][j];
34             ave2[j] = ave1[j] / n;//Averaging
35             if(j == m - 1)
36                 printf("%.2lf",ave2[j]);
37             else
38                 printf("%.2lf ",ave2[j]);
39         }
40         cout << endl;
41         //It is calculated that the scores of several students in all subjects are higher than the average
42         int temp,count = 0;
43         for(int i = 0;i < n;i++)//Sequential traversal
44         {
45             temp = 0;
46             for(int j = 0;j < m;j++)
47                 if(fenshu[i][j] >= ave2[j])//According to the question, if a student's score in a certain subject is greater than or equal to the average score of the subject, then the temporary variable temp add one-tenth
48                     temp++;
49             if(temp == m)//Explain if the student's m If the scores of each subject are greater than the average scores of each subject, then the conditions are met. Add one to the counter
50                 count++;
51         }
52         cout << count << endl << endl;//Pay attention to the output format
53     }
54     return 0;
55 }```

This is actually a water problem. It's a new one on the test platform of our school. I did it for 2-3 hours, and I cried/~~

The idea is that the input student's score can be regarded as a matrix, and only the rows and columns of the matrix need to be processed and judged. It's too much for me/~~

Tags: C++ PHP

Posted on Tue, 05 Nov 2019 09:55:21 -0500 by jesus_hairdo