Given an array arr[] containing N distances of inch-feet system, such that each element of the array represents a distance in the form of {inch, feet}. The task is to add all the N inch-feet distances using structures.
Examples:
Input: arr[] = { { 10, 3.7 }, { 10, 5.5 }, { 6, 8.0 } };
Output:
Feet Sum: 27
Inch Sum: 5.20Input: arr[] = { { 1, 1.7 }, { 1, 1.5 }, { 6, 8 } };
Output:
Feet Sum: 8
Inch Sum: 11.20
Approach:
- Traverse the struct array arr and find the summation of all the inches of the given set of N distances as:
feet_sum = feet_sum + arr[i].feet; inch_sum = inch_sum + arr[i].inch;
- If the sum of all the inches (say inch_sum) is greater than 12, then convert the inch_sum into feet because
1 feet = 12 inches
Therefore update inch_sum to inch_sum % 12. Then find the summation of all the feets(say feet_sum) of N distances and add inches_sum/12 to this sum.
- Print the feet_sum and inch_sum individually.
Below is the implementation of the above approach:
// C program for the above approach #include "stdio.h" // Struct defined for the inch-feet system struct InchFeet {
// Variable to store the inch-feet
int feet;
float inch;
}; // Function to find the sum of all N // set of Inch Feet distances void findSum( struct InchFeet arr[], int N)
{ // Variable to store sum
int feet_sum = 0;
float inch_sum = 0.0;
int x;
// Traverse the InchFeet array
for ( int i = 0; i < N; i++) {
// Find the total sum of
// feet and inch
feet_sum += arr[i].feet;
inch_sum += arr[i].inch;
}
// If inch sum is greater than 11
// convert it into feet
// as 1 feet = 12 inch
if (inch_sum >= 12) {
// Find integral part of inch_sum
x = ( int )inch_sum;
// Delete the integral part x
inch_sum -= x;
// Add x%12 to inch_sum
inch_sum += x % 12;
// Add x/12 to feet_sum
feet_sum += x / 12;
}
// Print the corresponding sum of
// feet_sum and inch_sum
printf ( "Feet Sum: %d\n" , feet_sum);
printf ( "Inch Sum: %.2f" , inch_sum);
} // Driver Code int main()
{ // Given set of inch-feet
struct InchFeet arr[]
= { { 10, 3.7 },
{ 10, 5.5 },
{ 6, 8.0 } };
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
findSum(arr, N);
return 0;
} |
// C++ program for the above approach #include "iostream" using namespace std;
// Struct defined for the inch-feet system struct InchFeet {
// Variable to store the inch-feet
int feet;
float inch;
}; // Function to find the sum of all N // set of Inch Feet distances void findSum(InchFeet arr[], int N)
{ // Variable to store sum
int feet_sum = 0;
float inch_sum = 0.0;
int x;
// Traverse the InchFeet array
for ( int i = 0; i < N; i++) {
// Find the total sum of
// feet and inch
feet_sum += arr[i].feet;
inch_sum += arr[i].inch;
}
// If inch sum is greater than 11
if (inch_sum >= 12) {
// Find integral part of inch_sum
int x = ( int )inch_sum;
// Delete the integral part x
inch_sum -= x;
// Add x%12 to inch_sum
inch_sum += x % 12;
// Add x/12 to feet_sum
feet_sum += x / 12;
}
// Print the corresponding sum of
// feet_sum and inch_sum
cout << "Feet Sum: "
<< feet_sum << '\n'
<< "Inch Sum: "
<< inch_sum << endl;
} // Driver Code int main()
{ // Given a set of inch-feet
InchFeet arr[]
= { { 10, 3.7 },
{ 10, 5.5 },
{ 6, 8.0 } };
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
findSum(arr, N);
return 0;
} |
Feet Sum: 27 Inch Sum: 5.20
Time Complexity: O(N), where N is the number inch-feet distances.