In the contests interface, I can filter those contests that I have already attempted, but there is no such feature in gyms. What should I do in this case? Is there any method for that?

102222I - Bubble Sort

2018-2019 ACM-ICPC, China Multi-Provincial Collegiate Programming Contest

We define $$$A$$$ is the permutation of n, and $$$B$$$ is a sequence of n.

$$$B_i$$$ is defined as the number of subscripts that satisfy the condition

Obviously,after a round of bubbling,$$$B_i$$$will minus one (if $$$B_i>0$$$), and $$$A_i$$$ will also move forward one space (i--)(although this is useless).

We consider to use sequence $$$B$$$ to solve the problem, we first prove that for every valid sequence $$$B$$$, there is exactly one permutation $$$A$$$ corresponding to it.

We consider how to infer $$$A$$$ from $$$B$$$.

It is very obvious that $$$A_n$$$ (the length of the permutation is n) can be directly infered by $$$B_n$$$, and $$$A_{n-1}$$$ is also easily to know if $$$A_n$$$ is known,and so on. So as long as $$$B$$$ is legal, there is an $$$A$$$ permutation obtained, certificated.

That is to say, for each bubbling, every $$$B_i$$$ greater than $$$0$$$ is decremented by one, and equal to $$$0$$$, it remains unchanged.

The transformation of $$$B$$$ is easy to stimulate by the conclusion (Obviously,...).

So we only need to calculate the legal number of sequence of $$$B$$$.

Consider what final states are legal:

1. All $$$B_i=0$$$ (ie $$$A$$$ is ordered)

2. There is a consecutive subsequence of $$$B_i=1$$$ in the permutation, and the rest are 0, such as 0 0 1 1 0 (1 4 2 3 5)

3. There is only one $$$B_i!=0$$$ in the permutation, and the others are 0, such as 0 0 2 0 0 (2 3 1 4 5)

I think it is easy to use Dynamic programming to solve this problem, set 3 states is ok.

        memset(dp,0,sizeof(dp));
        int ans2=0;
        cin>>n>>m>>mod;
        dp[0][0]=1;
        for(int i=1;i<=n;i++){
            if(i<=m+1){
                (dp[i][0]+=dp[i-1][0]*(i))%=mod;
            }
            else{
                (dp[i][0]+=dp[i-1][0]*(m+1))%=mod;
                (dp[i][1]+=dp[i-1][0]+dp[i-1][1])%=mod;
                (dp[i][2]+=dp[i-1][1]*(m+1)+dp[i-1][2]*(m+1)+dp[i-1][0]*max(0ll,i-1-(m+1)))%=mod;
            }
        }
        int ans=0;
        for(int i=0;i<3;i++)ans+=dp[n][i];
        cout<<"Case #"<<ts<<": "<<ans%mod<<"\n";