Codeforces Round #175 (Div. 2) C. Building Permutation（贪心）

C. Building Permutation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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).

思路：对于若a[i]满足0<a[i]<=n，则标记tag[a[i]]=1,否则tag[a[i]]=1.对于集合A={a[i]|a[i]<=0 || a[i]>n}，进行不降序排列，集合B=[1,n]-｛a[i]|0<a[i]<=n｝也不降序排列，A中元素转化为B中位置对应相同的元素。

AC Code：

#include <iostream>
#include <fstream>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <utility>
using namespace std;
#define ll long long
#define cti const int
#define ctll const long long
#define dg(i) cout << '*' << i << endl;

int p[300002], pi;
bool tag[300002];

int main()
{
    int n, x;
    ll cnt;
    while(scanf("%d", &n) != EOF)
    {
        memset(tag, false, sizeof(tag));
        pi = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &x);
            if(x < 1 || x > n || tag[x]) p[pi++] = x;
            else tag[x] = true;
        }
        sort(p, p + pi);
        cnt = 0;
        for(int i = 0, j = 1; i < pi; i++, j++)
        {
            while(tag[j]) j++;
            cnt += (ll)abs(p[i] - j);
        }
        printf("%I64d\n", cnt);
    }
    return 0;
}