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
题意：
给出n个数的数组，从数组中找到一个最长子串，子串中只能有k个不同的数字，输出子串的第一个元素的下标和最后一个元素的下标。数组n的下标从1开始计数。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int vis[6000007]; //vis数组来标记这个数字是否出现过
int a[600007];
int main()
{
int n, k;
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; ++ i)
{
scanf("%d", &a[i]);
}
memset(vis,0,sizeof(vis));
int l=1,r=1;//存放最终子串的左边界和右边界
int num=0;//存放子串中不同的数的个数
int ll=1;//代表左指针
for(int i=1;i<=n;i++) //i来代表右指针
{
if(vis[a[i]]==0) //如果这个数没出现过
{
vis[a[i]]++;
num++;
}
else //这个数字已经存在过
{
vis[a[i]]++;
}
while(num>k) //如果不相同的数的个数大于k
{
vis[a[ll]]--; //左指针就要向右移动，直到num≤k
if(vis[a[ll]]==0)
num--;
ll++;
}
if((i-ll+1)>(r-l+1)) //不断更新子串中的数字个数
{
r=i;
l=ll;
}
}
printf("%d %d\n",l,r);
return 0;
}