C. Serval and Toxel's Arrays

链接：https://www.luogu.com.cn/problem/CF1789C or https://codeforces.com/problemset/problem/1789/C
题目：

大佬思路：

有点像链式前向星的思路，就是通过记录前一个的位置来记录连续存储长度。
代码：

#define _CRT_SECURE_NO_WARNINGS 
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
#define IOS ios::sync_with_stdio(false), cin.tie(0) ,cout.tie(0)
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
int lst[N*2], sum[N*2],a[N];
int n, m;
 
int main()
{
	IOS;
	int t; cin >> t;
	while (t--)
	{
		cin >> n >> m;
		memset(lst, -1, sizeof(lst)), memset(sum, 0, sizeof(sum));
		int maxn = 0;
		for (int i = 0; i < n; i++) { int x; cin >> x; lst[x] = 0; a[i + 1] = x; maxn = max(maxn, x); }
		for (int i = 1; i <= m; i++)
		{
			int ax, bx; 
			cin >> ax >> bx;maxn = max(maxn, bx);
			sum[a[ax]] += i - lst[a[ax]];
			lst[a[ax]] = -1;
			a[ax] = bx;
			lst[bx] = i;
		}
		ll ans = 0;
		for (int i = 1; i <= n; i++)
			if (lst[a[i]] !=-1)
				sum[a[i]] += m + 1 - lst[a[i]];
		for (int i = 1; i <= maxn; i++)
		{
			ll x = sum[i];
			ans += x * (x - 1) / 2 + x * (m + 1 - x);
		}
		cout << ans << '\n';
	}
 
	return 0;
}

注意要开unsigned long long防越界