Educational Codeforces Round 6 C

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 n 给一段长为n的数字 问最多有多少段 每段要求：有两个相同的数字
题解 贪心处理 直接贪心会超时 set 处理 注意最后一段的右区间一定为n

2000ms 超时代码

#include<bits/stdc++.h>
using namespace std;
int n;
struct node
{
    int l;
    int r;
}m[300005];
int a[300005];
int main()
{
    scanf("%d",&n);
    int st=0;
    int jishu=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        for(int j=st+1;j<i;j++)
        {
            if(a[j]==a[i])
            {
                m[jishu].l=st+1;
                m[jishu].r=i;
                jishu++;
                st=i;
            }
        }
    }
    if(jishu==0)
        printf("-1\n");
    else
    {
        printf("%d\n",jishu);
        for(int i=0;i<jishu-1;i++)
        {
            printf("%d %d\n",m[i].l,m[i].r);
        }
        printf("%d %d\n",m[jishu-1].l,n);
    }
  return 0;
}

　　

set 处理  218ms

#include<bits/stdc++.h>
using namespace std;
struct node
{
    int l;
    int r;
}m[300005];
int exm;
set<int>s;
set<int>::iterator it;
int n;
int main()
{
    scanf("%d",&n);
    int jishu=0;
    for(int i=1;i<=n;)
    {
        m[jishu].l=i;
        while(i<=n)
        {
            scanf("%d",&exm);
            i++;
            it=s.find(exm);
            if(it!=s.end())
            {
                s.clear();
                m[jishu].r=i-1;
                jishu++;
                break;
            }
            else
                s.insert(exm);
        }
    }
    if(jishu==0)
        printf("-1\n");
    else
    {
        printf("%d\n",jishu);
        for(int i=0;i<jishu-1;i++)
        {
            printf("%d %d\n",m[i].l,m[i].r);
        }
        printf("%d %d\n",m[jishu-1].l,n);
    }
    return 0;
}