Codeforces Round #323 (Div. 2) D. Once Again ... Violence + Longest non-descending subsequence

D. Once Again ...

You is given an array of positive integers a1, a2, ..., an x t of length n x T. We know that for any i > n It was true that ai = a i - n. Find the length of the longest non-decreasing sequence of the given array.

Input

The first line contains the space-separated integers: n, T (1≤ n ≤100, 1≤ t ≤10 7). The second line contains n space-separated integers a1, a2, ..., C16>an (1≤ ai ≤300).

Output

Print a single number-the length of a sought sequence.

Sample Test (s) input

4 3
3 1 4 2

Output

5

Note

The array given in the sample looks like That:3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in Bol D form the largest non-decreasing subsequence.

Test instructions: The longest non-descending subsequence length for you to ask for n*t

The puzzle: Because it is the T-N arrangement, the middle segment must be the same, it must be the largest number of n permutations, in t less than 100 is the violent DP, 100 when the calculation of the rain is good

///1085422276#include <iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<cmath>#include<map>#include<bitset>#include<Set>#include<vector>using namespacestd; typedefLong Longll;#defineMem (a) memset (A,0,sizeof (a))#defineMeminf (a) memset (A,127,sizeof (a));#defineTS printf ("111111\n");#definefor (I,A,B) for (int i=a;i<=b;i++)#defineForj (I,A,B) for (int i=a;i>=b;i--)#defineREAD (a,b,c) scanf ("%d%d%d", &a,&b,&c)#defineMoD 1000000007#defineINF 100000inline ll read () {ll x=0, f=1; CharCh=GetChar ();  while(ch<'0'|| Ch>'9')    {        if(ch=='-') f=-1; CH=GetChar (); }     while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; CH=GetChar (); }    returnx*F;}//****************************************#defineMAXN 100+5intA[MAXN];intdp[30005];inthashs[ -];intMain () {intn=read (); intt=read (); For (I,1, N) {scanf ("%d",&A[i]); } for (I,0N102) dp[i]=1; if(t<= -) {for (I,1, T) {for (J,1, N) {                 for(intk=1; k<j+n* (I-1); k++)                {                    inttmp=k%N; if(tmp==0) tmp=N; if(a[j]>=a[tmp]) dp[j+n* (I-1)]=max (dp[j+n* (i-1)],dp[k]+1); }            }        }        intmm=-1;  for(intI=1; i<=n*t;i++) mm=Max (DP[I],MM); cout<<mm<<Endl; }    Else{for (I,1, -) {for (J,1, N) {                 for(intk=1; k<j+n* (I-1); k++)                {                    inttmp=k%N; if(tmp==0) tmp=N; if(a[j]>=a[tmp]) dp[j+n* (I-1)]=max (dp[j+n* (i-1)],dp[k]+1); }            }        }        intmm=-1, flag,ans=0;  for(intI=1; i<=n* -; i++) mm=Max (DP[I],MM);  for(intI=1; i<=n;i++) {Hashs[a[i]]++; if(Hashs[a[i]]>ans) ans=Hashs[a[i]]; } cout<<mm+ (t -) *ans<<Endl; }    return 0;}

Code

Codeforces Round #323 (Div. 2) D. Once Again ... Violence + Longest non-descending subsequence