求等距离收集物品所覆盖的距离
一场比赛正在进行，几块石头被放在一条路上。比赛起点放置一个水桶，距离第一块石头 5 个单位。其他的石头相距 3 个单位，一个接一个地排成一条直线。即第一石和第二石之间的距离是 3 个单位，第三石和第四石之间，也是 3 个单位，以此类推。竞争对手从水桶开始，捡起最近的石头，回去把那块石头放进水桶，然后再次跑去收集下一个最近的石头，跑回来，把它放进水桶。这样，这个过程一直持续到所有的石头都被放进桶里。 现在如果有 n 块石头躺在地上，那么在完成整个过程中，竞争对手会走多远。
示例:
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:
竞争对手摘第一块石头跑的距离= 2 * 5 竞争对手摘第二块石头跑的距离= 2(5 + 3) 竞争对手摘第三块石头跑的距离= 2(5 + 3 + 3) = 2(5 + (2 * 3)) 竞争对手摘第四块石头跑的距离= 2(5 + 3 + 3 + 3) = 2(5 + (3 * 3)) 竞争对手摘第五块石头跑的距离= 2(5 + 3 + 3 + 3 + 3) = 。 。 竞赛者摘第 n 块石头跑的距离= 2(5 + 3 + 3 + ……)。+ (n-1)次) = 2(5 + (n-1) 3) 所以选手跑的总距离=以上所有距离的总和 =(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 次)+(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 *
以下是实施办法:
C++
// C++ program to calculate
// the distance for given problem
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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));
}
}
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
//PHP program to calculate
// the distance for given problem
// function to calculate the
// distance
function find_distance($n)
{
return $n * ((3 * $n) + 7);
}
// Driver program
$n = 5;
echo "Distance = ", find_distance($n);
// This code is contributed by aj_36
?>
java 描述语言
<script>
// 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));
</script>
输出:
110
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