C. Building Permutation

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最新推荐文章于 2023-02-23 22:08:12 发布

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本文介绍了一个算法问题，即如何通过最少的加1或减1操作将一个整数序列转化为一个包含1到n的唯一整数的排列。文章提供了一种解决方案，包括对输入序列进行排序并计算每个元素与其目标位置之间的距离之和。

Permutation p is an ordered set of integers p₁, p₂, ..., p_n, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as p_i. We'll call number n the size or the length of permutationp₁, p₂, ..., p_n.

You have a sequence of integers a₁, a₂, ..., a_n. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.

Input

The first line contains integer n (1 ≤ n ≤ 3·10⁵) — the size of the sought permutation. The second line contains n integersa₁, a₂, ..., a_n ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

Print a single number — the minimum number of moves.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the%I64d specifier.

Sample test(s)

input
2
3 0

output
2

input
3
-1 -1 2

output
6

Note

In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).

In the second sample you need 6 moves to build permutation (1, 3, 2).

解题说明：此题的意思就是如何通过最小的加1减1操作把原数列改造成只含1-n每个数字只含一次的数列。既然是最小的步数，那么我们应该把原数列中的每个数字尽量往其最接近的1-n中的数字靠。为此需要进行排序，然后通过对差的绝对值求和即可。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

double a[300001];

int main()
{
	int n,i,j;
	double temp;
	double sum;
	sum=0;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%lf", &a[i]);
	}
	sort(a,a+n);
	for(i=0;i<n;i++)
	{
		sum+=abs(i+1-a[i]);
	}
	printf("%.0lf\n",sum);
	return 0;
}