Codeforces Round #323 (Div. 2) D. Once Again... 暴力+最长非递减子序列

                                                                              D. Once Again...

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

Input

The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., 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 bold form the largest non-decreasing subsequence.

题意：要你求n*T的最长非递减子序列长度

题解：由于是T个n排列，中间段必定是相同的，也必定是n排列中数最多的，在T小于100是暴力dp，大雨100时计算就好了

///
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define mod 1000000007
#define inf 100000
inline ll read()
{
    ll x=,f=;
    char ch=getchar();
    while(ch<''||ch>'')
    {
        if(ch=='-')f=-;
        ch=getchar();
    }
    while(ch>=''&&ch<='')
    {
        x=x*+ch-'';
        ch=getchar();
    }
    return x*f;
}
//****************************************

#define maxn 100+5
int a[maxn];
int dp[];
int hashs[];
int main()
{

    int n=read();
    int T=read();
    FOR(i,,n)
    {
        scanf("%d",&a[i]);
    }
    FOR(i,,n*)dp[i]=;
    if(T<=)
    {
        FOR(i,,T)
        {
            FOR(j,,n)
            {
                for(int k=;k<j+n*(i-);k++)
                {
                    int tmp=k%n;
                    if(tmp==)tmp=n;
                    if(a[j]>=a[tmp])
                        dp[j+n*(i-)]=max(dp[j+n*(i-)],dp[k]+);
                }
            }
        }
        int mm=-;
        for(int i=;i<=n*T;i++)mm=max(dp[i],mm);
        cout<<mm<<endl;
    }
    else {

       FOR(i,,)
        {
            FOR(j,,n)
            {
                for(int k=;k<j+n*(i-);k++)
                {
                    int tmp=k%n;
                    if(tmp==)tmp=n;
                    if(a[j]>=a[tmp])
                        dp[j+n*(i-)]=max(dp[j+n*(i-)],dp[k]+);
                }
            }
        }
        int mm=-,flag,ans=;
        for(int i=;i<=n*;i++)mm=max(dp[i],mm);
           for(int i=;i<=n;i++)
           {
               hashs[a[i]]++;
               if(hashs[a[i]]>ans)ans=hashs[a[i]];
           }
           cout<<mm+(T-)*ans<<endl;
    }
    return ;
}

代码