2021-3-23-meituan-regular sequence

Title Description ：
We call a length n The sequence of is a regular sequence , If and only if the sequence is a sequence composed of 1~n The arrangement of components , That is, the sequence consists of n It's made up of four positive integers , The value is [1,n] Range , And there are no duplicate numbers , At the same time, regular sequences do not require sorting
One day, the small group got a length of n Any sequence of , He needs to operate in a limited number of operations , Turn this sequence into a regular sequence , He can choose a number in the sequence at any time , And add or subtract one to the number .
How many operations can he use at least to turn this sequence into a regular sequence ？

Input description :
The first line of input contains only one positive integer n, Represents the length of any sequence .(1<=n<=20000)
The second line of input contains n It's an integer , Represents the given sequence , The absolute value of each number is less than 10000.

Output description :
The output contains only one integer , Indicates the minimum number of operations .

Example 1
Input
5
-1 2 3 10 100

Output
103

// Greedy Algorithm , We want to minimize the number of steps each number moves ,
// Or let each number find [1,n] Nearest pit filling 
// We can sort first , Then press all the numbers [1,n] The pit is filled from small to large 

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main()
{
    
	int n;
	cin >> n;
	vector<int> vecNums;

	int nTemp = 0;
	for (int i = 0; i < n; i++)
	{
    
		cin >> nTemp;
		vecNums.push_back(nTemp);
	}

	sort(vecNums.begin(), vecNums.end());

	int k = 1;
	int ans = 0;
	for (int i = 0; i < n; i++)
	{
    
		ans += abs(k - vecNums[i]);
		k++;
	}
	cout << ans << endl;
	
	return 0;
}