Codeforces Round 923 (Div. 3) D. Find the Different Ones!

写点简单的思维题
https://codeforces.com/problemset/problem/1927/D

思路：用两个数组，一个存储原始数据，一个用nex存该位置第一次不一样的下标

#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>
typedef long long ll;
using namespace std;


int nex[200010];
int alst[200010];
int main()
{
	int t; cin >> t; 
	
	int nn = 0;
	for (int iii = 0; iii < t; iii++)
	{
		memset(nex, 0, sizeof(nex));
		memset(alst, 0, sizeof(alst));
		nn = 0;
		int n; cin >> n;
		for (int i = 0; i < n; i++) {
			cin >> alst[i];
			if (i > 0 and alst[i] != alst[i - 1])
			{
				nex[nn] = i;
				nn = i;
			}
		}
		for (int i = 1; i < n; i++)
		{
			if (alst[i] == alst[i - 1])
				nex[i] = nex[i - 1];
		}
		int tt; cin >> tt;
		for (int ii = 0; ii < tt; ii++)
		{
			int l, r;
			cin >> l >> r;
			l--, r--;
			if (nex[l] > r or nex[l] == 0)cout << "-1 -1" << endl;//这里不能落了nex[l] == 0的判断，因为还有可能根本就不存在这个数，当然也可以把都是0的拿length替换，思路不变
			else cout << nex[l]+1 << ' ' << l+1 << endl;//注意这里的下标
		}
		cout << endl;
	}
	return 0;
}