Write a program that finds the average time spent programming by a student each

Question

Write a program that finds the average time spent programming by a student each day over a 3 day period. Hints and what program should include: Must use Nested for Loops. Prompt the user for the number of students. Prompt the user for the time spent programming by each student for each of the 3 days, this should be iin hours Print to the screen the average number of hours per day spent programming by each student. Label the output to include the student number. Output Run: 4 Students. Student 1 (5, 5, 5); Student 2 (8 2 1 Student 3 (4 25 15 Student 4(3 7 0)

Explanation / Answer

#include<iostream>

using namespace std;

int main()

{

int n;

//reading number of students

cout<<"Enter number of students:";

cin>>n;

//declaring array;

double student[n];

int i=1;

while(i<=n)//loop

{

int j=1;

double x,s=0;

cout<<" Student "<<i<<endl;

while(j<=3)//nested loop

{

//reading hours

cout<<" Time spent programming on day "<<j<<"(in hours) :"<<endl;

cin>>x;

s=s+x;

j++;

}

student[i-1]=s/3;//computing average

i++;

}

//displaying output

i=1;

cout<<" Student Average_Time_spent ";

while(i<=n)

{

//displaying output

cout<<"Student "<<i<<" "<<student[i-1]<<endl;

i++;

}

return 0;

}

output:

Enter number of students:4

Student 1
Time spent programming on day 1(in hours) :
5
Time spent programming on day 2(in hours) :
5
Time spent programming on day 3(in hours) :
5

Student 2
Time spent programming on day 1(in hours) :
8
Time spent programming on day 2(in hours) :
2
Time spent programming on day 3(in hours) :
1

Student 3
Time spent programming on day 1(in hours) :
4
Time spent programming on day 2(in hours) :
2.5
Time spent programming on day 3(in hours) :
1.5

Student 4
Time spent programming on day 1(in hours) :
3
Time spent programming on day 2(in hours) :
7
Time spent programming on day 3(in hours) :
0

Student Average_Time_spent
Student 1 5
Student 2 3.66667
Student 3 2.66667
Student 4 3.33333

Process exited normally.
Press any key to continue . . .

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