Gym100956 Problem J. Sort It!

https://codeforces.com/group/NVaJtLaLjS/contest/253990/attachments

思路：
設$len(i,j):$以$i$爲結尾，長度爲j的合法子序列個數。那麼$f(i,j)=\sum_{k=1}^{i-1}f(k,j-1)$，用樹狀數組優化。（$n^3 -> n^2*log(n)$）注意樹狀數組的sum函數裏面要加取模！
$lensum(i)爲長度爲i的合法子序列個數$
設$cnt(i):i$個不同的字母拼成n個的組合數目。那麼由容斥原理得$cnt(i)=i^n-\sum_{j=1}^{i-1}C_i^j*cnt(j)$
答案就是$\sum_{i=1}^n lensum(i)*cnt(i)$

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lowbit(x) (x&(-x))
const int maxn=2050;
const int mod=1000000007;

ll n,a[maxn],len[maxn][maxn],lensum[maxn],c[maxn],cnt[maxn],C[maxn][maxn];

int sum(int x)
{
    ll ret=0;
    while(x)ret=(ret+c[x])%mod,x-=lowbit(x);
    return ret;
}

void add(int x,int d)
{
    while(x<=n)c[x]+=d,x+=lowbit(x);
}

ll pow_mod(ll a,ll n)
{
    if(!n)return 1;
    ll x=pow_mod(a,n/2);
    x=x*x%mod;
    if(n&1)x=x*a%mod;
    return x;
}

int main()
{
    //freopen("input.in","r",stdin);
    cin>>n;
    for(int i=0;i<=n;i++)
    {
        C[i][0]=1;
        for(int j=1;j<=i;j++)C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
    }
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<=n;i++)len[a[i]][1]=1;
    for(int j=2;j<=n;j++)
    {
        for(int i=1;i<=n;i++)
        {
            len[a[i]][j]=sum(a[i]-1);
            add(a[i],len[a[i]][j-1]);
        }
        memset(c,0,sizeof(c));
    }
    for(int j=1;j<=n;j++)for(int i=1;i<=n;i++)lensum[j]=(lensum[j]+len[i][j])%mod;
    for(int i=1;i<=n;i++)
    {
        cnt[i]=pow_mod(i,n);
        for(int j=i-1;j>0;j--)cnt[i]=(cnt[i]-cnt[j]*C[i][j]%mod+mod)%mod;
    }
    ll ans=0;
    for(int i=1;i<=n;i++)ans=(ans+cnt[i]*lensum[i])%mod;
    cout<<ans<<endl;
    return 0;
}