找出要收集等距离物品的覆盖距离

一场比赛正在进行，在道路上放置了几块石头。在比赛的起点放置一个水桶，与第一块石头相距5个单位。其他石头彼此相距3个单位，并排成一条直线。即，第一和第二个石头之间的距离是3个单位，第三和第四个石头之间的距离也是3个单位，依此类推。竞争对手从水桶开始，捡起最近的石头，然后回去将其放入桶中，然后再次奔跑以收集下一个最近的石头，向后跑，然后将其放入桶中。这样，过程将继续进行，直到将所有石头都放入铲斗中为止。
现在，如果地面上有n块石头，那么在完成整个过程中，竞争对手将覆盖多少距离。

例子：

Input : n = 3
Output : Distance = 48
Explanation
= 2*5 + 2(5 + 3) + 2(5 + 3 + 3)
= 10 + 16 + 22
= 48

Input : n = 5
Output : Distance = 110
Explanation
= 2*5 + 2(5 + 3) + 2(5 + 3 + 3) + 2(5 + 3 + 3 + 3) + 2(5 + 3 + 3 + 3 + 3)
= 10 + 16 + 22 + 28 + 34
= 110

观察模式：

Distance run by competitor to pick first stone = 2 * 5 
Distance run by competitor to pick second stone = 2(5 + 3)
Distance run by competitor to pick third stone = 2(5 + 3 + 3) 
= 2(5 + (2 * 3))
Distance run by competitor to pick fourth stone = 2(5 + 3 + 3 + 3) 
= 2(5 + (3 * 3))
Distance run by competitor to pick fifth stone = 2(5 + 3 + 3 + 3 + 3) 
= 2(5 + (4 * 3)) 
. 
. 
.
Distance run by competitor to pick n-th stone = 2(5 + 3 + 3 + ……. + (n-1) times ) 
= 2(5 + (n-1) *3)
So total distance run by competitor = sum of all the above distances 
= (2 * 5) + 2(5 + 3) + 2(5 + (2 * 3)) + 2(5 + (3 * 3)) + ………….. + 2(5 + (n-1) *3) 
= 2(5 + (5 + 3) + (5 + (2 * 3)) + (5 + (3 * 3)) + ………………. + (5 + (n-1) * 3) 
= 2(5 + 5 + 5 …… + n times) + (3 + (2 * 3) + (3 * 3) + ……… + (n-1) * 3) 
= 2(5n + 3(1 + 2 + 3 + ……………. + n-1)) 
= 2(5n + 3/2[(n-1)*(n-1 + 1)] ) 
= 2(5n + 3/2[(n-1)*n]) 
= 2(5n + 3/2(n² – n)) 
= 10n + 3*n² – 3*n 
= 3*n² + 7*n 
= n*((3 * n) + 7)

下面是该方法的实现：

C++

// C++ program to calculate
// the distance for given problem
#include 
using namespace std;
 
// function to calculate the
// distance
int find_distance(int n)
{
    return n * ((3 * n) + 7);
}
 
// Driver program
int main()
{
    int n = 5;
    cout << "Distance = " << find_distance(n);
    return 0;
}

Java

// Java program to calculate the
// distance for given problem
class demo {
 
    // function to calculate
    // the distance
    public static int find_distance(int n)
    {
        return n * (3 * n + 7);
    }
 
    // Driver program
    public static void main(String args[])
    {
        int n = 5;
 
        System.out.print("Distance = ");
        System.out.println(find_distance(n));
    }
}

Python3

# Python3 code to calculate
# the distance for given problem
 
# function to calculate the
# distance
def find_distance(n):
     
    return n * ((3 * n) + 7)
 
# main function
n = 5
ans = find_distance( n )
print (ans)
 
# This code is contributed by Saloni Gupta

C#

// C# program to calculate
// the distance for given problem
using System;
 
class GFG {
 
    // function to calculate the
    // distance
    public static int find_distance(int n)
    {
        return n * ((3 * n) + 7);
    }
 
    // Driver program
    public static void Main()
    {
        int n = 5;
 
        Console.Write(find_distance(n));
    }
}
 
// This code is contributed by
// Smitha Dinesh Semwal

Javascript

 
// JavaScript program to calculate the
// distance for given problem
 
// function to calculate
    // the distance
    function find_distance(n)
    {
        return n * (3 * n + 7);
    }
 
// Driver code
 
        let n = 5;
   
        document.write("Distance = ");
        document.write(find_distance(n));
 

输出：

110