查找给定圆的两个部分的最小角度差的程序

给定一个将圆形划分为n个大小为n的数组的数组。数组的第i个元素表示一件的角度。我们的任务是由这些零件制成两个连续的零件，以使这两个零件的角度之差最小。
例子 ：

Input : arr[] = {90, 90, 90, 90}
Output : 0
In this example, we can take 1 and 2 
pieces and 3 and 4 pieces. Then the 
answer is |(90 + 90) - (90 + 90)| = 0.

Input : arr[] = {170, 30, 150, 10}
Output : 0
In this example, we can take 1 and 4, 
and 2 and 3 pieces. So the answer is 
|(170 + 10) - (30 + 150)| = 0.

Input : arr[] = {100, 100, 160}
Output : 40

我们可以注意到，如果其中一个零件是连续的，那么所有其余零件也将形成一个连续的零件。如果第一部分的角度等于x，则第一部分和第二部分的角度之差为| x –(360 – x)|。 = | 2 * x – 360 | = 2 * | x – 180 |。因此，对于每个可能的连续部分，我们都可以计算其角度并更新答案。

C++

// CPP program to find minimum
// difference of angles of two
// parts of given circle.
#include 
using namespace std;
 
// Returns the minimum difference
// of angles.
int findMinimumAngle(int arr[], int n)
{
    int l = 0, sum = 0, ans = 360;
    for (int i = 0; i < n; i++) {
        // sum of array
        sum += arr[i];
 
        while (sum >= 180) {
 
            // calculating the difference of
            // angles and take minimum of
            // previous and newly calculated
            ans = min(ans, 2 * abs(180 - sum));
            sum -= arr[l];
            l++;
        }
 
        ans = min(ans, 2 * abs(180 - sum));
    }
    return ans;
}
 
// driver code
int main()
{
    int arr[] = { 100, 100, 160 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findMinimumAngle(arr, n) << endl;
    return 0;
}
 
// This code is contributed by "Abhishek Sharma 44"

Java

// java program to find minimum
// difference of angles of two
// parts of given circle.
import java.util.*;
 
class Count{
    public static int findMinimumAngle(int arr[], int n)
    {
        int l = 0, sum = 0, ans = 360;
        for (int i = 0; i < n; i++)
        {
            // sum of array
            sum += arr[i];
     
            while (sum >= 180)
            {
     
                // calculating the difference of
                // angles and take minimum of
                // previous and newly calculated
                ans = Math.min(ans,
                            2 * Math.abs(180 - sum));
                sum -= arr[l];
                l++;
            }
     
            ans = Math.min(ans,
                            2 * Math.abs(180 - sum));
        }
         
        return ans;
         
    }
     
    public static void main(String[] args)
    {
        int arr[] = { 100, 100, 160 };
        int n = 3;
        System.out.print(findMinimumAngle(arr, n));
    }
}
 
// This code is contributed by rishabh_jain

Python3

# java program to find minimum
# difference of angles of two
# parts of given circle.
import math
 
# function returns the minimum
# difference of angles.
def findMinimumAngle (arr, n):
    l = 0
    _sum = 0
    ans = 360
    for i in range(n):
         
        #sum of array
        _sum += arr[i]
         
        while _sum >= 180:
         
            # calculating the difference of
            # angles and take minimum of
            # previous and newly calculated
            ans = min(ans, 2 * abs(180 - _sum))
            _sum -= arr[l]
            l+=1
        ans = min(ans, 2 * abs(180 - _sum))
    return ans
     
# driver code
arr = [100, 100, 160]
n = len(arr)
print(findMinimumAngle (arr, n))
 
# This code is contributed by "Abhishek Sharma 44"

C#

// C# program to find minimum
// difference of angles of two
// parts of given circle.
using System;
 
class GFG
{
    public static int findMinimumAngle(int []arr, int n)
    {
        int l = 0, sum = 0, ans = 360;
        for (int i = 0; i < n; i++)
        {
            // sum of array
            sum += arr[i];
     
            while (sum >= 180)
            {
     
                // calculating the difference of
                // angles and take minimum of
                // previous and newly calculated
                ans = Math.Min(ans,
                      2 * Math.Abs(180 - sum));
                sum -= arr[l];
                l++;
            }
     
            ans = Math.Min(ans,
                        2 * Math.Abs(180 - sum));
        }
         
        return ans;
         
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = { 100, 100, 160 };
        int n = 3;
        Console.WriteLine(findMinimumAngle(arr, n));
    }
}
 
// This code is contributed by vt_m

PHP

= 180)
        {
 
            // calculating the difference of
            // angles and take minimum of
            // previous and newly calculated
            $ans = min($ans, 2 *
                             abs(180 - $sum));
            $sum -= $arr[$l];
            $l++;
        }
 
        $ans = min($ans, 2 * abs(180 - $sum));
    }
    return $ans;
}
 
// Driver Code
$arr = array( 100, 100, 160 );
$n = sizeof($arr);
echo findMinimumAngle($arr, $n), "\n" ;
 
// This code is contributed by m_kit
?>

Javascript

// javascript program to find minimum
// difference of angles of two
// parts of given circle.
  
let MAXN = 109;
   
    function findMinimumAngle(arr, n)
    {
        let l = 0, sum = 0, ans = 360;
        for (let i = 0; i < n; i++)
        {
            // sum of array
            sum += arr[i];
       
            while (sum >= 180)
            {
       
                // calculating the difference of
                // angles and take minimum of
                // previous and newly calculated
                ans = Math.min(ans,
                            2 * Math.abs(180 - sum));
                sum -= arr[l];
                l++;
            }
       
            ans = Math.min(ans,
                            2 * Math.abs(180 - sum));
        }
           
        return ans;
           
    }
     
// Driver code
 
        let arr = [ 100, 100, 160 ];
        let n = 3;
        document.write(findMinimumAngle(arr, n));
 

输出 ：

40