Educational Codeforces Round 6 C. Pearls in a Row （分离序列+set）

C. Pearls in a Row

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n pearls in a row. Let's enumerate them with integers from 1 to n from the left to the right. The pearl number i has the type a_i.

Let's call a sequence of consecutive pearls a segment. Let's call a segment good if it contains two pearls of the same type.

Split the row of the pearls to the maximal number of good segments. Note that each pearl should appear in exactly one segment of the partition.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains integer n (1 ≤ n ≤ 3·10⁵) — the number of pearls in a row.

The second line contains n integers a_i (1 ≤ a_i ≤ 10⁹) – the type of the i-th pearl.

Output

On the first line print integer k — the maximal number of segments in a partition of the row.

Each of the next k lines should contain two integers l_j, r_j (1 ≤ l_j ≤ r_j ≤ n) — the number of the leftmost and the rightmost pearls in the j-th segment.

Note you should print the correct partition of the row of the pearls, so each pearl should be in exactly one segment and all segments should contain two pearls of the same type.

If there are several optimal solutions print any of them. You can print the segments in any order.

If there are no correct partitions of the row print the number "-1".

Sample test(s)

Input

5
1 2 3 4 1

Output

1
1 5

Input

5
1 2 3 4 5

Output

-1

Input

7
1 2 1 3 1 2 1

Output
2
1 3
4 7


题意：给你一串含有n个数的序列，然后让你分离这个序列，要求连续且其中最多的数不能多于2个

思路：如果数据小的话，数组就可以了，但是数据是1e9，所以说用set容器。。。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
struct s
{
	int l,r;
}ans[MAXN];
set<ll>q;
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		q.clear();
		ll num;
		int cnt=0;
		int begin=1;
		for(int i=1;i<=n;i++)
		{
			scanf("%I64d",&num);
			if(q.count(num))
			{
				ans[cnt].l=begin;ans[cnt].r=i;cnt++;
				q.clear();
				begin=i+1;
			}
			else
			q.insert(num);
		}
		if(ans[cnt-1].r<n)
		ans[cnt-1].r=n;
		if(cnt>0)
		{
			printf("%d\n",cnt);
			for(int i=0;i<cnt;i++)
			printf("%d %d\n",ans[i].l,ans[i].r);
			
		}
		else
		printf("-1\n");
	}
	return 0;
}