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
这道题可以采用尺取法,可以将本题的时间复杂度降低到O( n),尺取法, 就是使用类似于卡尺的方法一步步移动区间范围(即左右端点),解决连续区间的一些问题。
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[600000];
int vis[6000000];
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 num = 0, pos = 1, maxs = 0;
int l = 1, r = k;
for(int i = 1; i <= n; ++ i)
{
if(vis[a[i]] == 0)
num++;
vis[a[i]] ++;
while(num > k)
{
vis[a[pos]]--;
if(vis[a[pos]] == 0)
num--;
pos++;
}
if(maxs < (i - pos + 1))
{
maxs = i - pos + 1;
l = pos;
r = i;
}
}
cout << l << ' ' << r << endl;
return 0;
}