P4137

dp #数学 #计数dp #组合

从大到小考虑将每个数加入当前的序列

\(dp_{i,j}\) 表示 \([1,i]\) 全部加入后，\(LISNumber=j\) 的方案数

对于一个新的数，有 \(x\) 个，枚举这个数加入的时候分为几段 \(k\) ，再枚举其中有多少段 \(cnt\) 加在一个极大递增序列后面

那么 \(LISNumber\) 的变化量就是

\[x-k+(k-cnt) \]

达到这个的方案数为

\[\binom{x-1}{k}\binom{j}{cnt}\binom{sum-j+1}{x-cnt} \]

其中 \(sum=\sum\limits_{j=1}^{i-1}a_j\)

直接暴力转移

// Author: xiaruize
#ifndef ONLINE_JUDGE
bool start_of_memory_use;
#endif
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
clock_t start_clock = clock();
#endif
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
    if (a > b)
        return a;
    return b;
}
int min(int a, int b)
{
    if (a < b)
        return a;
    return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

int c[2005][2005];
void init()
{
    c[0][0] = 1;
    rep(i, 1, 2000)
    {
        c[i][0] = c[i][i] = 1;
        rep(j, 1, i - 1)
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
    }
}

int getc(int x, int y)
{
    if (x < 0 || y < 0 || x < y)
        return 0;
    if (x == y || y == 0)
        return 1;
    return c[x][y];
}

int n, len;
int a[2005];
int sum = 0;
int dp[40][2005];

void solve()
{
    init();
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    cin >> len;
    dp[0][0] = 1;
    rep(i, 1, n)
    {
        rep(j, 0, len)
        {
            int x = a[i];
            rep(k, 1, x)
            {
                rep(cnt, 0, k)
                {
                    int rev = k - cnt;
                    int tmp = getc(x - 1, k - 1);
                    int p = x - k;
                    (dp[i][j + p + rev] += dp[i - 1][j] * tmp % MOD * getc(j, cnt) % MOD * getc(sum - j + 1, rev) % MOD) %= MOD;
                    // (dp[i][j + p + 1] += dp[i - 1][j] * tmp % MOD * getc(j, k - 1) % MOD) %= MOD;
                }
            }
        }
        sum += a[i];
        // rep(j, 0, len) cerr << dp[i][j] << ' ';
        // cerr << endl;
    }
    cout << dp[n][len] << endl;
}

#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    // cin >> testcase;
    while (testcase--)
        solve();
#ifndef ONLINE_JUDGE
    cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
    cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
    return 0;
}