题解 | #美妙的约会#

参考解法https://blog.csdn.net/meng_lemon/article/details/97141897

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;

int main()
{
    int n;
    cin >> n;
    int nums[2*n];
    for(int i=0; i<2*n; i++)
    {
        cin >> nums[i];
    }

    int count = 0, left = 0, right;
    for(left; left<2*n; left=left+2)
    {
        right = left + 1;
        //找到值相等的对应元素的下标
        while(right<2*n && nums[left]!=nums[right])    right++;
        //将对应元素与前一个元素交换位置，直到下标为left+1，即值相等两元素相邻
        for(right; right>left+1; right--)
        {
            swap(nums[right-1], nums[right]);
            count ++;
        }
    }
    cout << count <<endl;
    return 0;
}