C/C++程序,使用结构
添加英寸英尺系统中给定的 N 个距离
原文:https://www . geesforgeks . org/c-CPP-program-to-add-n-distance-给定英寸-英尺-系统使用结构/
给定一个包含英寸-英尺系统的 N 距离的数组 arr[] ,使得数组的每个元素以{英寸,英尺} 的形式表示一个距离。任务是使用结构添加所有 N 英寸英尺距离。
示例:
输入: arr[] = { { 10,3.7 },{ 10,5.5 },{ 6,8.0 } }; T3】输出:T5】尺和:27 寸和:5.20
输入: arr[] = { { 1,1.7 },{ 1,1.5 },{ 6,8 } }; T3】输出:T5】尺和:8 寸和:11.20
进场:
-
遍历结构数组 arr ,求给定的一组 N 距离的所有英寸的总和,如下所示:
``` 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
```
因此将英寸 _ 总和更新为英寸 _ 总和% 12 。然后求 N 距离的所有英尺(比如英尺 _ 总和)的总和,并将英寸 _ 总和/12 加到这个总和上。
-
分别打印英尺 _ 总和和英寸 _ 总和。
下面是上述方法的实现:
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
时间复杂度: O(N) ,其中 N 为英寸-英尺距离数。
版权属于:月萌API www.moonapi.com,转载请注明出处