/* ELEN 1301 Programming Assignment #5. Name : Your name. Student ID : Your student ID #. Due date : Due date Purpose of the pro
gram : Finding the average, minimum and maximum values of seven entered numbers. Use iomanip library to show 3 digits below decimal point. Section 1 : Enter the first number. Set min and max variables to the entered number. Section 2 : Enter the next number. If it is smaller than min, replace the min with the entered number. If it is bigger than max, replace the max with the entered number. Section 3 : Repeat section 2 five more times, so that you have seven numbers. Section 4 : Calculate the average of the seven numbers and show the result with 3 digits below decimal point. (2 points) Section 5 : Show the minimum number. (2 points) Section 6 : Show the maximum number. (2 points) */ #include #include using namespace std; int main() { int n1, n2, n3, n4, n5, n6, n7, min, max; double sum = 0; // Write the rest of the program. return 0; } // main
1 answer:
Answer:
Here is the C++ program:
#include<iostream> //to use input output functions
#include<iomanip> //to use setprecision
using namespace std; //to identify objects cin cout
int main() { //start of main function
int n1, n2, n3, n4, n5, n6, n7, min, max; // declare variables for 7 numbers, minimum value and maximum value
double sum= 0; //declare variable to hold sum of 7 numbers
double average; //declare variable to hold average of 7 numbers
cout<<"Enter first number: "; //prompts user to enter 1st number
cin>>n1; //reads first number from user
max = n1; //sets the first number to maximum
min=n1; //sets the first number to minimum
cout<<"Enter second number: "; //prompts user to enter 2nd number
cin>>n2; //reads second number from user
if(n2<min){ //if second number is less than min
min=n2; } //sets min to n2
if(n2>max){ //if n2 is greater than max
max = n2; } //sets max to n2
cout<<"Enter third number: "; //prompts user to enter 3rd number
cin>>n3; //reads third number from user
if(n3<min){ //checks if n3 is greater than min
min=n3; } //sets n3 to min
if(n3>max){ //checks if n3 is greater than max
max = n3; } //sets max to n3
cout<<"Enter fourth number: ";//prompts user to enter 4th number
cin>>n4; //reads fourth number from user
if(n4<min){ //if n4 is less than min
min=n4; } //sets min to n4
if(n4>max){ //if n4 is greater than max
max = n4; } //sets max to n4
cout<<"Enter fifth number: "; //prompts user to enter 5th number
cin>>n5; //reads fifth number from user
if(n5<min){ //if n5 is less than min
min=n5; } //sets min to n5
if(n5>max){ //if n5 is greater than max
max = n5; } //sets max to n5
cout<<"Enter sixth number: "; //prompts user to enter 6th number
cin>>n6; //reads sixth number from user
if(n6<min){ // if n6 is less than min
min=n6; } //sets min to n6
if(n6>max){ //if n6 is greater than max
max = n6; } //sets max to n6
cout<<"Enter seventh number: ";//prompts user to enter 7th number
cin>>n7; //reads seventh number from user
if(n7<min){ //if n7 is less than minimum number
min=n7; } //assigns n7 to min
if(n7>max){ //if n7 is greater than the maximum number
max = n7; } //assigns n7 to max
sum = n1+n2+n3+n4+n5+n6+n7; //adds 7 numbers
average = sum/7; //computes average of 7 numbers
cout<<"The average is: "<<fixed<<setprecision(3)<<average<<endl; //displays average value up to show 3 digits below decimal point using setprecision method of iomanip library
cout<<"The maximum number is: "<<max<<endl; //displays maximum number of 7 numbers
cout<<"The minimum number is: "<<min<<endl; //displays miimum number of 7 numbers
return 0; }
Explanation:
The program is well explained in the comments attached to each statement of program. For example if
n1 = 3
n2 = 9
n3 = 7
n4 = 6
n5 = 2
n6 = 5
n7 = 4
When n1 is read using cin then this number is set to max and min as:
min = 9
max = 3
Now when n2 is read, the first if condition checks if n2 is less than min and second if condition checks if n2 is greater than max. As n2 = 9 so it is not less than min so this if condition is false and n2 is greater than max i.e. 3 so this condition is true. So 9 is assigned to max.
min = 9
max = 9
Now when n3 is read the values of min and max become:
min = 7
max = 9
Now when n4 is read the values of min and max become:
min = 6
max = 9
Now when n5 is read the values of min and max become:
min = 2
max = 9
Now when n6 is read the values of min and max become:
min = 2
max = 9
Now when n7 is read the values of min and max become:
min = 2
max = 9
Now the statement sum = n1+n2+n3+n4+n5+n6+n7;
executes which becomes:
sum = 3 + 9 + 7 + 6 + 2 + 5 + 4
sum = 36.0
Next program control moves to statement:
average = sum/7;
this becomes
average = 36/7
5.142857
Since this is to be displayed up to 3 decimal places so average = 5.143
the complete output of the program is attached.
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