使圆弧段相等大小所需的最小切割数

给定一个数组$N$元素，其中数组中的每个元素表示圆上已有切口的度(0 <= a [i] <= 359) 。任务是找到使圆弧段大小相等所需的最小额外切割次数。
例子：

Input : arr[] = { 0, 2 }
Output : 178

Input : arr[] = { 30, 60, 180 }
Output : 9

方法：解决上述问题的一种有效方法是找到所有连续的角度差的gcd。该gcd是一个圆形线段的最大角度，然后将获得360 / gcdOb的线段数。但是，已经有N个削减。因此，其他削减将为(360 / gcdObtained)– N。
下面是上述方法的实现：

C++

// CPP program to find the minimum number
// of additional cuts required to make
// circle segments equal sized
 
#include 
using namespace std;
 
// Function to find the minimum number
// of additional cuts required to make
// circle segments are equal sized
int minimumCuts(int a[], int n)
{
    // Sort the array
    sort(a, a + n);
 
    // Initial gcd value
    int gcd = a[1] - a[0];
    int s = gcd;
 
    for (int i = 2; i < n; i++) {
        gcd = __gcd(gcd, a[i] - a[i - 1]);
        s += a[i] - a[i - 1];
    }
 
    // Inlcuding the last segment
    if (360 - s > 0)
        gcd = __gcd(gcd, 360 - s);
 
    return (360 / gcd) - n;
}
 
// Driver code
int main()
{
    int arr[] = { 30, 60, 180 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << minimumCuts(arr, n);
 
    return 0;
}

Java

// Java program to find the minimum
// number of additional cuts required
// to make circle segments equal sized
import java.util.Arrays;
 
class GFG
{
     
// Recursive function to
// return gcd of two nos
static int findgcd(int a, int b)
{
    if (b == 0)
        return a;
    return findgcd(b, a % b);
}
 
// Function to find the minimum number
// of additional cuts required to make
// circle segments are equal sized
static int minimumCuts(int a[], int n)
{
    // Sort the array
    Arrays.sort(a);
 
    // Initial gcd value
    int gcd = a[1] - a[0];
    int s = gcd;
 
    for (int i = 2; i < n; i++)
    {
        gcd = findgcd(gcd, a[i] - a[i - 1]);
        s += a[i] - a[i - 1];
    }
 
    // Inlcuding the last segment
    if (360 - s > 0)
        gcd = findgcd(gcd, 360 - s);
 
    return (360 / gcd) - n;
}
 
// Driver code
public static void main(String[] args)
{
    int[] arr = new int[] { 30, 60, 180 };
 
    int n = arr.length;
 
    System.out.println(minimumCuts(arr, n));
}
}
 
// This code is contributed by mits

Python 3

# Python 3 program to find the minimum number
# of additional cuts required to make
# circle segments equal sized
  
import math
# Function to find the minimum number
# of additional cuts required to make
# circle segments are equal sized
def minimumCuts(a, n):
     
    # Sort the array
    a.sort()
  
    # Initial gcd value
    gcd = a[1] - a[0]
    s = gcd
  
    for i in range(2,n) :
        gcd = math.gcd(gcd, a[i] - a[i - 1])
        s += a[i] - a[i - 1]
  
    # Inlcuding the last segment
    if (360 - s > 0):
        gcd = math.gcd(gcd, 360 - s)
  
    return (360 // gcd) - n
  
# Driver code
if __name__ == "__main__":
    arr = [ 30, 60, 180 ]
  
    n = len(arr)
  
    print(minimumCuts(arr, n))

C#

// C# program to find the minimum
// number of additional cuts required
// to make circle segments equal sized
 
using System;
class GFG
{
// Recursive function to
// return gcd of two nos
static int findgcd(int a, int b)
{
if (b == 0)
return a;
 
return findgcd(b, a % b);
}
 
// Function to find the minimum number
// of additional cuts required to make
// circle segments are equal sized
static int minimumCuts(int []a, int n)
{
// Sort the array
Array.Sort(a);
 
// Initial gcd value
int gcd = a[1] - a[0];
int s = gcd;
 
for (int i = 2; i < n; i++)
{
gcd = findgcd(gcd, a[i] - a[i - 1]);
s += a[i] - a[i - 1];
}
 
// Inlcuding the last segment
if (360 - s > 0)
gcd = findgcd(gcd, 360 - s);
 
return (360 / gcd) - n;
}
 
// Driver Code
static void Main()
{
int[] arr = new int[] { 30, 60, 180 };
int n = arr.Length;
 
Console.WriteLine(minimumCuts(arr, n));
}
// This code is contributed by ANKITRAI1
}

PHP

 0)
        $gcd = findgcd($gcd, 360 - $s);
     
    return (360 / $gcd) - $n;
}
 
// Driver Code
$arr = array(30, 60, 180);
$n = sizeof($arr);
 
echo (minimumCuts($arr, $n));
 
// This code is contributed by ajit
?>

Javascript

// javascript program to find the minimum
// number of additional cuts required
// to make circle segments equal sized
 
    // Recursive function to
    // return gcd of two nos
    function findgcd(a , b)
    {
        if (b == 0)
            return a;
        return findgcd(b, a % b);
    }
 
    // Function to find the minimum number
    // of additional cuts required to make
    // circle segments are equal sized
    function minimumCuts(a, n)
    {
     
        // Sort the array
        a.sort();
 
        // Initial gcd value
        var gcd = a[1] - a[0];
        var s = gcd;
 
        for (i = 2; i < n; i++) {
            gcd = findgcd(gcd, a[i] - a[i - 1]);
            s += a[i] - a[i - 1];
        }
 
        // Inlcuding the last segment
        if (360 - s > 0)
            gcd = findgcd(gcd, 360 - s);
 
        return (360 / gcd) - n;
    }
 
    // Driver code
        var arr = [ 30, 60, 180 ];
        var n = arr.length;
        document.write(minimumCuts(arr, n));
 
// This code is contributed by aashish1995

输出：

9