Question

1. Using Dev C++, write a program to solve the problem as stated below.
2. Write a program will input a single integer value (all input will be exactly 8 digits long) and will store pairs of digits as individual integer values. The program will display the five 2-digit numbers as well as the average of these, each on their own line. Use a switch statement to print out Grade: and then either A (>=90), B ([80..89]), C ([70..79]), D ([60..69]) or F (<60) based on the calculated average.
   1. Sample
      1. Input: 54861398
      2. Output:       

54

86

13

98

Avg: 62.75

Grade: D

2. Sample
   1. Input: 64536785
   2. Output:        

64

92

67

85

Avg: 77.00

Grade: D

3. make use of a repetition structure (counter-controlled while loop) to shorten part of your code.
4. Compile the program and correct errors.

Answer 1

Program:

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main()
{
long number,num1,num2;
int sum=0,i,arr[4];
double avg;
printf("Enter 8 digit number: ");
scanf("%ld",&number);

//code to get 4 pairs
num1=number/10000;
num2=number%10000;

i=0;
while(i<=3)
{
if(i==0)
{
arr[i]=num1/100;
}
else if(i==1)
{
arr[i]=num1%100;
}
else if(i==2)
{
arr[i]=num2/100;
}
else if(i==3)
{
arr[i]=num2%100;
}
i++;
}
printf("\n4 pairs of %ld number: \n\n",number);
for(i=0;i<=3;i++)
{
printf("%d\n",arr[i]);
sum=sum+arr[i]; //sum of all pairs
}

avg=(double) sum/4; //average

printf("\nAvg: %.2f",avg);

switch((int)avg/10) //getting grade
{
case 10:
case 9:
printf("\nGrade: A\n");
break;
case 8:
printf("\nGrade: B\n");
break;
case 7:
printf("\nGrade: C\n");
break;
case 6:
printf("\nGrade: D\n");
break;
default:
printf("\nGrade: F\n");
  
}

return 0;
}

Output: