CF995B Suit and Tie 贪心 第十三

Suit and Tie

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Allen is hosting a formal dinner party. 2n2n people come to the event in nn pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The 2n2n people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this makes the picture more aesthetic.

Help Allen find the minimum number of swaps of adjacent positions he must perform to make it so that each couple occupies adjacent positions in the line.

Input

The first line contains a single integer nn (1≤n≤1001≤n≤100), the number of pairs of people.

The second line contains 2n2n integers a1,a2,…,a2na1,a2,…,a2n. For each ii with 1≤i≤n1≤i≤n, ii appears exactly twice. If aj=ak=iaj=ak=i, that means that the jj-th and kk-th people in the line form a couple.

Output

Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions.

Examples

input

Copy

4
1 1 2 3 3 2 4 4

output

Copy

2

input

Copy

3
1 1 2 2 3 3

output

Copy

0

input

Copy

3
3 1 2 3 1 2

output

Copy

3

Note

In the first sample case, we can transform 11233244→11232344→1122334411233244→11232344→11223344 in two steps. Note that the sequence 11233244→11323244→1133224411233244→11323244→11332244 also works in the same number of steps.

The second sample case already satisfies the constraints; therefore we need 00 swaps.

题意：给你2*n个数，问你把相同的数放在一起最小需要移动多少次位置

一个简单的贪心，找每次和自己相同的数的位置，中间已经被找到的数记为-1，不可访问。

然后加上每次找到的位置减去数自己的位置再减去中间已经被找到的数再减去一

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e3 + ;
const int mod = 1e9 + ;
typedef long long ll;
int main(){
    std::ios::sync_with_stdio(false);
    ll n;
    while( cin >> n ) {
        ll a[maxn], sum = ;
        for( ll i = ; i < *n; i ++ ) {
            cin >> a[i];
        }
        for( ll i = ; i < *n; i ++ ) {
            if( a[i] != - ) {
                ll num = ;
                for( ll j = i + ; j < *n; j ++ ) {
                    if( a[j] == - ) {
                        num ++;
                    }
                    if( a[j] == a[i] ) {
                        sum += ( j - i - num );
                        a[j] = -;
                        break;
                    }
                }
            }
        }
        cout << sum << endl;
    }
    return ;
}