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-goodif 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 nfrom 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
AC代码(尺取法维护一个符合要求的区间)
Select Code
#include <iostream>
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int s[500030],vis[1000030];
int main()
{
int n,m;
while (scanf("%d%d",&n,&m)!=EOF)
{
for (int i=0;i<n;i++)
scanf("%d",&s[i]);
memset(vis,0,sizeof(vis));
int pos=0,maxx=0,ansl=0,ansr=0,num=0;
for (int i=0;i<n;i++)
{
if (vis[s[i]]==0)
num++;
vis[s[i]]++;
while (num>m)
{
vis[s[pos]]--;
if (vis[s[pos]]==0) num--;
pos++;
}
if (maxx<i-pos+1)
{
maxx=i-pos+1;
ansl=pos;
ansr=i;
}
}
printf("%d %d\n",ansl+1,ansr+1);
}
}