Farmer John's N cows (1 <= N
<= 100,000) are lined up in a row. Each cow is identified by an
integer "breed ID" in the range 0...1,000,000,000; the breed ID of the
ith cow in the lineup is B(i). Multiple cows can share the same breed
ID. FJ thinks that his line of cows will look much more impressive if
there is a large contiguous block of cows that all have the same breed
ID. In order to create such a block, FJ chooses up to K breed IDs and
removes from his lineup all the cows having those IDs. Please help FJ
figure out the length of the largest consecutive block of cows with the
same breed ID that he can create by doing this.

* Line 1: Two space-separated integers: N and K.
* Lines 2..1+N: Line i+1 contains the breed ID B(i).

* Line 1: The largest size of a contiguous block of cows with identical breed IDs that FJ can create.

INPUT DETAILS: There are 9 cows in the lineup, with breed IDs 2, 7, 3, 7, 7, 3, 7, 5, 7. FJ would like to remove up to 1 breed ID from this lineup.

OUTPUT DETAILS: By removing all cows with breed ID 3, the lineup reduces to 2, 7, 7, 7, 7, 5, 7. In this new lineup, there is a contiguous block of 4 cows with the same breed ID (7).

Gold

这题刚开始不会，后来才知道在一个序列中，必须是连续，然后只有k+1种数，然后找出从中的

出现过的数的次数的最大值即可，用map维护，每次直接更新当前加入后该元素个数，不需要找最大值

因为在答案中，绝对是以其结尾出现过的。

 #pragma GCC optimize(2)

 #pragma G++ optimize(2)

 #include<iostream>

 #include<algorithm>

 #include<cmath>

 #include<cstdio>

 #include<cstring>

 #include<map>

 #define N 10000007

 using namespace std;

 inline int read()

 {

     int x=,f=;char ch=getchar();

     while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}

     while(isdigit(ch)){x=(x<<)+(x<<)+ch-'';ch=getchar();}

     return x*f;

 }

 int n,k,ans;

 int a[N];

 map<int,int>p;

 int main()

 {

     n=read(),k=read()+;

     for (int i=;i<=n;i++)a[i]=read();

     int num=,mx=,l=;

     for (int i=;i<=n;i++)

     {

         if(!p[a[i]])num++;

         p[a[i]]++;

         while(num>k)

         {

             if(p[a[l]]==)p.erase(a[l]),num--;

             else p[a[l]]--;

             l++;

         }

         ans=max(ans,p[a[i]]);

 //        cout<<i<<" "<<ans<<endl;

     }

     printf("%d\n",ans);

 }