codeforces 285C Building Permutation (简单贪心)

Building Permutation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Permutationp is an ordered set of integersp₁,  p₂,  ...,  p_n, consisting ofn distinct positive integers, each of them doesn't exceedn. We'll denote thei-th element of permutation p asp_i. We'll call numbern 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 containsn integers a₁, 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.

Examples

Input

23 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).

问题：就是给你一组数...然后让它变为自然数数列（不用管顺序）所需要的最小改变...

分析：问题其实就是找每个数和自然数列差的绝对值的和最小...而自然数列是递增数列...所以把题目里给的数列排成递增的，然后算下差的绝对值的和就行...

注意：有个坑是用来记改变的sum值用int的话会越界...所以还是得用long long...其他的都用int就行...注意题目中给的范围...

#include<iostream>#include<algorithm>using namespace std;int main(){int n,a[300005];cin>>n;for(int i=0;i<n;i++){cin>>a[i];}sort(a,a+n);int t=1;long long sum=0;for(int j=0;j<n;j++){sum+=abs(j+1-a[j]);}cout<<sum<<endl;return 0;}