Educational Codeforces Round 5 （D. Longest k-Good Segment）

D. Longest k-Good Segment

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The array a withn integers is given. Let's call the sequence of one or more consecutive elements inasegment. Also let's call the segmentk-good if it contains no more thank different values.

Find any longest k-good segment.

As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to usescanf/printf instead ofcin/cout in C++, prefer to useBufferedReader/PrintWriter instead ofScanner/System.out inJava.

Input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·10⁵) — the number of elements in a and the parameterk.

The second line contains n integersa_i (0 ≤ a_i ≤ 10⁶) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from1 ton from left to right.

Sample test(s)

Input

5 5
1 2 3 4 5

Output

1 5

Input

9 3
6 5 1 2 3 2 1 4 5

Output

3 7

Input

3 1
1 2 3

Output

1 1

题意：第一行给你两个数m,k;

第二行给你n个数;

要求你找一个连续区间，这个区间不同的数的个数不超过k而且这个区间长度最大，输出区间端点l,r（任意一组就行）

分析：two pointers解决，知道这个方法做这个题简直就是水题，直接看代码。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[500010];
int vis[1000010];
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)==2)
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        memset(vis,0,sizeof(vis));
        int len=1,l=1,r=1,sum=1;///len记录最长的长度，用来更新答案而已，l和r分别指向区间两端，sum表示的是a[l]到a[r]中不同的数的个数
        int x=1,y=1;
        vis[a[1]]=1;///预处理第一个值
        while(r<=n)
        {
            r++;
            if(r>n) break;
            if(vis[a[r]]==0)///加入这个值之前没有这个值，那么不同的数多了1，sum++
            {
                sum++;
                vis[a[r]]++;
            }
            else vis[a[r]]++;
            while(sum>k)
            {
                vis[a[l]]--;
                if(!vis[a[l]]) sum--;///如果这个值-1后这个值就没有了，代表少了一个不同的数，sum--
                l++;
            }
            if(r-l+1>len)
            {
                len=r-l+1;
                x=l,y=r;
            }
        }
        printf("%d %d\n",x,y);
    }
    return 0;
}