102222I - Bubble Sort

Obviously, if there are Bi numbers k that satisfy the conditionk<i&&Ak>Ai, then after a round of bubbling,the number of kAk>Ai&&k>i will minus one, and Ai will also move forward one space i--(although this is useless)

We consider Bi to solve the problem, we first prove that for every valid B, there is exactly one A corresponding to it

We consider how to infer A from B

It is very obvious that An can be directly infered by Bn, and A{n-1} is also easily to know if An is known, so as long as B is legal, there is an A permutation obtained, certificated

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

Consider what final states are legal:

1. All Bi=0 (ie ordered)

2. There is a Bi=1 in the middle, the rest are 0, such as 0 0 1 1 0 (1 4 2 3 5)

3. There is a Bi!=0 in the middle, 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 (all m in the front, in the middle section, all m in the back)

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";