The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k 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 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 two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.
The second line contains n integers ai (0 ≤ ai ≤ 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 from 1 to n from left to right.
Examples
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
题目要求：从所给序列找到一段尽量长的连续系列，使得这段序列不同的数字个数不大于K。
（因为英语不行，连题目意思都理解错了。。。）
题目要求从所给序列找到一段尽量长的连续系列，使得这段序列不同数字个数(cnt)不大于K。
一个暴力的思路是：对于原始序列枚举每个点作为左端点，找到对应最大的右端点。记为(Li,Ri)
假设从左向右边枚举，当枚举Li+1时，先处理Li位置，然后，明显当前Ri<=Ri+1;从Ri+1开始修改cnt值就可以了。
最后取最大值（Ri-Li）。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;
int a[500010];
int v[1000010];//根据题意ai的大小开的数组大小；
int main()
{
int n, k, i;
scanf("%d%d", &n, &k);
for(i=1; i<=n; i++)
{
scanf("%d", &a[i]);
}
memset(v, 0, sizeof(v));
int j=1;//类似于栈，栈的底部;
int l=1, r=1, num=0;
for(i=1; i<=n; i++)
{
v[a[i]]++;
if(v[a[i]]==1)
num++;//记录这段序列不同的数字个数;
while(num>k)//
{
v[a[j]]--;//当这段序列不同的数字个数大于k时，栈后移。
if(v[a[j]]==0)
num--;//因为题目中可以允许a[i]的值可以重复，所以只有v[a[]]为零时，才能减一。
j++;
}
if(r-l < i-j)
{
l = j;
r = i;
}
}
printf("%d %d\n", l, r);
return 0;
}