环游中回到起点的最少步骤

考虑圆形轨道，其中n个点标记为1、2，… n。一个人最初放置在点k上。人移动m> 0，在每一步中向前移动(循环)。找到使人到达初始点k所需的最小步数。
例子：

Input : n = 9, k = 2, m = 6 
Output : 3
Explanation : Sequence of moves is
 2 => 8 => 5 => 2

Input : n = 6, k = 3, m = 2 
Output : 3

天真的方法：用“ k”和“ count” = 0初始化计数器“ i”。此外，对于每个迭代增量，将“ count”添加“ m”到“ i”。如果i> n，则取n为模，即i =((i + m)％n) 。如果我等于k，那么计数将是我们的答案。
时间复杂度： O(n)。
高效方法：我们找到GCD(n，m)，然后将n除以GCD(n，m)。那将是我们的答案。可以解释为：
现在考虑每个问题的n和m，因为我们知道gcd(n，m)必须除以n，并且商告诉我们，从起始位置(例如0)开始连续m次跳跃(加)m个数字之后，我们再次到达起始位置。
注意：在n个数字的圆形排列中，第n个位置和第0个位置相同。

C++

// C++ program to find minimum steps to reach
// starting position in a circular tour.
#include
using namespace std;
 
// function for finding minimum steps
int minStroke(int n, int m)
{
    // return value n / gcd(n, m)
    return (n/__gcd(n, m));
}
 
// Driver function
int main()
{
    int n = 12, k = 5, m = 8;
    cout << minStroke(n, m);
    return 0;
}

Java

// Java program to find minimum steps to reach
// starting position in a circular tour.
 
class Test
{
    // method for finding minimum steps
    static int minStroke(int n, int m)
    {
        // return value n / gcd(n, m)
        return (n/gcd(n, m));
    }
     
    // method for gcd
    static int gcd(int n, int m) {
          
        if (n == 0 || m == 0)
           return 0;
       
         
        if (n == m)
            return n;
       
        if (n > m)
            return gcd(n-m, m);
        return gcd(n, m-n);
    }
 
    // Driver method
    public static void main(String args[])
    {
        int n = 12, k = 5, m = 8;
        System.out.println(minStroke(n, m));
    }
}

Python3

# Python program to find minimum
# steps to reach starting position
# in a circular tour.
 
# function for finding minimum steps
def minStroke(n, m):
     
    # return value n / gcd(n, m)
    return (n / __gcd(n, m))
 
# method for gcd
def __gcd(n, m):
     
    if(n == 0 or m == 0):
        return 0
         
    if (n == m):
        return n
         
    if (n > m):
        return __gcd(n-m, m)
     
    return __gcd(n, m-n)
 
# Driver code
n = 12
k = 5
m = 8
 
print(minStroke(n, m))
 
# This code is contributed by Anant Agarwal.

C#

// C# program to find minimum steps to reach
// starting position in a circular tour.
using System;
using System.Collections;
 
class GFG
{
    // method for finding minimum steps
    static int minStroke(int n, int m)
    {
        // return value n / gcd(n, m)
        return (n/gcd(n, m));
    }
     
    // method for gcd
    static int gcd(int n, int m) {
         
        if (n == 0 || m == 0)
        return 0;
         
        if (n == m)
            return n;
         
        if (n > m)
            return gcd(n-m, m);
             
        return gcd(n, m-n);
    }
 
    // Driver method
    public static void Main()
    {
        int n = 12, m = 8;
        Console.WriteLine(minStroke(n, m));
    }
}
 
// This code is contributed by Sam007

PHP

 $b)
        return __gcd($a - $b , $b);
 
    return __gcd($a , $b - $a);
}
 
// function for
// finding minimum steps
function minStroke($n, $m)
{
    // return value n / gcd(n, m)
    return ($n / __gcd($n, $m));
}
 
// Driver Code
$n = 12; $k = 5; $m = 8;
echo minStroke($n, $m);
 
// This code is contributed by anuj_67.
?>

Javascript

 
// JavaScript program to find minimum steps to reach
// starting position in a circular tour.
 
// function for finding minimum steps
function minStroke(n, m)
{
    // return value n / gcd(n, m)
    return (n/gcd(n, m));
}
 
// method for gcd
function gcd(n, m) {
            
        if (n == 0 || m == 0)
           return 0;
         
           
        if (n == m)
            return n;
         
        if (n > m)
            return gcd(n-m, m);
        return gcd(n, m-n);
}
// Driver function
 
    let n = 12, k = 5, m = 8;
    document.write(minStroke(n, m));
     
// This code is contributed by Surbhi Tyagi.
 

输出：

3