CodeForces - 224B Array

Array

You’ve got an array a, consisting of n integers: a1, a2, …, an. Your task is to find a minimal by inclusion segment [l, r] (1 ≤ l ≤ r ≤ n) such, that among numbers al,  al + 1,  …,  ar there are exactly k distinct numbers.

Segment [l, r] (1 ≤ l ≤ r ≤ n; l, r are integers) of length m = r - l + 1, satisfying the given property, is called minimal by inclusion, if there is no segment [x, y] satisfying the property and less then m in length, such that 1 ≤ l ≤ x ≤ y ≤ r ≤ n. Note that the segment [l, r] doesn’t have to be minimal in length among all segments, satisfying the given property.

Input
The first line contains two space-separated integers: n and k (1 ≤ n, k ≤ 105). The second line contains n space-separated integers a1, a2, …, an — elements of the array a (1 ≤ ai ≤ 105).

Output
Print a space-separated pair of integers l and r (1 ≤ l ≤ r ≤ n) such, that the segment [l, r] is the answer to the problem. If the sought segment does not exist, print “-1 -1” without the quotes. If there are multiple correct answers, print any of them.

Examples
Input
4 2
1 2 2 3
Output
1 2
Input
8 3
1 1 2 2 3 3 4 5
Output
2 5
Input
7 4
4 7 7 4 7 4 7
Output
-1 -1
Note
In the first sample among numbers a1 and a2 there are exactly two distinct numbers.

In the second sample segment [2, 5] is a minimal by inclusion segment with three distinct numbers, but it is not minimal in length among such segments.

In the third sample there is no segment with four distinct numbers.
英语真是个好东西，这个题至少理解了5种意思。。。题意是输入一个长度为n的数组，再输入k，找到一个长度尽量短的区间，使得这个区间里有k个不同的数字。并且这个区间要尽量靠左。

#include<iostream>
#include<cmath>
#include<set>
using namespace std;
set<int>s,t;
int main()
{
    int a[100005];
    int n,i,m,j,k;
    bool flag,Flag;
    cin>>n>>m;
    for(i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    flag=false;Flag=false;
    for(i=1;i<=n;i++)
    {
        t.insert(a[i]);
        if(t.size()==m)
        {
            flag=true;
            j=i;
            break;
        }
    }
    for(i=j;i>=1;i--)
    {
        s.insert(a[i]);
        if(s.size()==m)
        {
            Flag=true;
            k=i;
            break;
        }
    }
    if(flag==true&&Flag==true)
    cout<<k<<" "<<j<<endl;
    else cout<<-1<<" "<<-1<<endl;
}