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

时间：2023-03-09 22:42:44
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

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