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
// 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++
// 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; } |
Output:
Feet Sum: 27 Inch Sum: 5.20
Time Complexity: O(N), where N is the number inch-feet distances.
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