CodeForces 285C

最新推荐文章于 2019-04-04 21:10:00 发布

原创最新推荐文章于 2019-04-04 21:10:00 发布 · 222 阅读

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本文介绍了一个经典的编程问题——如何通过增减操作将给定整数序列转换为排列，并提供了一种有效的解决方案。该方案首先对输入序列进行排序，然后计算序列元素与理想位置之间的绝对差值之和作为最小操作次数。

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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 permutation p₁, 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 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 C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

Copy

2
3 0

Output

Copy

Input

Copy

3
-1 -1 2

Output

Copy

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).

这道题只需要将输入的数组排序一下，并与原数组的差相加起来便是答案

#include<bits/stdc++.h>
using namespace std;
long long a[300005];
int main(){
	int n;
	while(~scanf("%d",&n)){
		
		for(int i=1;i<=n;i++){
			scanf("%I64d",&a[i]);
		}
		sort(a+1,a+1+n);
		__int64 ans=0;
		for(int i=1;i<=n;i++){
			ans+=abs(a[i]-i);
		}
		cout<<ans<<endl;
			
	}
}