code forces 996D Suit and Tie

D. 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  如何让一对一对匹配成功

1和1 匹配 2和2 匹配。。。（总感觉在虐狗)

只能两两交换位置移动

题解

从第一个开始找是否匹配，如果不匹配就从前往后找，找到后‘那一段’往后挪一个单位

代码如下

#include<bits/stdc++.h>
using namespace std;
int a[];
int main(){
    int n;
    while(~scanf("%d",&n)){
        for(int i=;i<*n;i++){
            scanf("%d",&a[i]);
        }
        int ans=;
        int pos;
        for(int i=;i<*n;i+=){
            if(a[i]!=a[i-]){
                int t=a[i];
               for(int j=i+;j<*n;j++){
                    if(a[j]==a[i-]){
                        ans+=j-i;
                        pos=j;
                        a[i]=a[j];
                        break;
                    }
               }
               //这个就是那一段
               for(int j=pos;j>i;j--){
                    a[j]=a[j-];
               }
                a[i+]=t;

            }
//            for(int j=0;j<2*n;j++){
//                printf("%d ",a[j]);
//            }
//            printf("\n");

        }
        printf("%d\n",ans);
    }
    return ;
}