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
思路：做了这几个尺取法的题目，总结出这样的一种套路，对于这种类型的尺取来说，这也可以当做一个模板了（和我上一篇博客差别不大）。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=5e5+100;
int a[maxx];
int n,k;
int main()
{
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
map<int,int> mp;
mp.clear();
int sum=0,l=1,r=1,_max=0;
int L=1,R=1;
while(r<=n)
{
while(sum<=k&&r<=n)
{
if(mp[a[r]]==0) sum++;
mp[a[r]]++;
r++;
}
if(sum>k) r--,mp[a[r]]--,sum--;
if(_max<r-l)
{
_max=r-l;
L=l,R=r-1;
}
while(sum>=k&&l<=n)
{
mp[a[l]]--;
if(mp[a[l]]==0) sum--;
l++;
}
}
printf("%d %d\n",L,R);
return 0;
}
努力加油a啊，(o)/~
