Educational Codeforces Round 6 620C Pearls in a Row(stl)

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 ai.

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/printfinstead 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 ai (1 ≤ ai ≤ 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 lj, rj (1 ≤ lj ≤ rj ≤ 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

題目鏈接:點擊打開鏈接

給出一個序列, 一段子序列含有兩個相同的數字稱爲happy segment, 要求分爲最多的happy segment, 輸出每段序列起點終點.

考慮存儲數據的數據結構, 使用set解決此問題, 遇到重複的元素則存儲位置且清空set, 同時要滿足第一段序列的起點爲1且最後一段序

列的終點爲n 即可.

AC 代碼:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 3e5 + 5;
int n, pos[MAXN], k;
set<int> s;
int main(int argc, char const *argv[])
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		int x;
		scanf("%d", &x);
		if(s.count(x) == 0) s.insert(x);
		else {
			pos[k++] = i;
			s.clear();
		}
	}
	if(k == 0) printf("-1\n");
	else {
		pos[k - 1] = n;
		printf("%d\n", k);
		printf("1 %d\n", pos[0]);
		for(int i = 0; i < k - 1; ++i)
			printf("%d %d\n", pos[i] + 1, pos[i + 1]);
	}
	return 0;
}