UOJ#273. 【清华集训2016】你的生命已如风中残烛

链接：

link

题解：

首先考虑如果确定了非零数的相对顺序怎么做。

考虑从左到右给每个wi贪心地分配离他最近的wi−1个0，那么我们记f(i,j)表示当前考虑分配wi，已经用了前j个空隙（数与数之间的位置），我们有转移：

f(i,j)=∑f(i−1,k)×gi,j−max(k,i)，其中gi,j=(wi−2+jj)。

g的含义就是将wi−1个0分配到j+1个空位，同时我们保证是极长的，所以最后一个0不参与分配的方案数。

这样就可以用一个状压DP来求方案数了。

注意到最后答案是一堆g乘起来，我们枚举n的自然数拆分，做两次DP，记pi表示对于i这种自然数拆分，所有g的积之和（不考虑顺序）；记qi表示这种拆分数中这样的顺序的方案数，那么ans=∑pi×qi。

代码：

#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){    int x = 0, f = 1, c = getchar();    for (; !isdigit(c); c = getchar())        if (c == '-')            f = -1;    for (;  isdigit(c); c = getchar())        x = x * 10 + c - '0';    return x * f;}const int MAXN = 200005;const int mod = 998244353;int n, m, nxt[45], sum[MAXN], fac[MAXN], inv[MAXN], way[MAXN], f[45][MAXN], trans[MAXN][45];map <vector <int>, int> idx;vector <int> cur, val[MAXN];inline int C(int n, int m){    return 1LL * fac[n] * inv[m] % mod * inv[n - m] % mod;}inline void Dfs(int s, int v, int r){    if (!s)        idx[cur] = ++ m, sum[m] = r, val[m] = cur;    else        for (int i = 1; i <= v && i <= s; i ++)            cur.pb(i), Dfs(s - i, i, r), cur.pop_back();}int main(){#ifdef wxh010910    freopen("data.in", "r", stdin);#endif    n = Read(), fac[0] = fac[1] = inv[0] = inv[1] = 1;    for (int i = 2; i < MAXN; i ++)        fac[i] = 1LL * fac[i - 1] * i % mod, inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;    for (int i = 2; i < MAXN; i ++)        inv[i] = 1LL * inv[i - 1] * inv[i] % mod;    for (int i = 0; i < n; i ++)        Dfs(i, i, i);    for (int i = 1; i <= m; i ++)    {        trans[i][0] = i;        for (int j = 1; j <= n - 1 - sum[i]; j ++)        {            vector <int> tmp = val[i];            tmp.pb(j);            sort(tmp.begin(), tmp.end(), greater <int> ());            trans[i][j] = idx[tmp];        }    }    f[0][1] = 1;    for (int i = 0; i < n; i ++)        for (int j = 1; j <= m; j ++)            if (f[i][j])                for (int k = 0; k <= i - sum[j]; k ++)                    f[i + 1][trans[j][k]] = (f[i + 1][trans[j][k]] + f[i][j]) % mod;    for (int i = 1; i <= m; i ++)        way[i] = f[n][i];    mset(f, 0);    f[0][1] = 1;    int ans = 0, cnt = 0;    for (int i = 0; i < n; i ++)    {        int x = Read() - 1;        cnt += x;        if (x)            for (int j = 0; j < n; j ++)                nxt[j] = C(x + j - 1, j);        else            for (int j = 0; j < n; j ++)                nxt[j] = !j;        for (int j = 1; j <= m; j ++)            if (f[i][j])                for (int k = 0; k <= n - 1 - sum[j]; k ++)                    f[i + 1][trans[j][k]] = (1LL * f[i][j] * nxt[k] + f[i + 1][trans[j][k]]) % mod;    }    for (int i = 1; i <= m; i ++)    {        int cur = 1LL * way[i] * f[n][i] % mod;        for (int l = 0, r = 0; l < val[i].size(); cur = 1LL * cur * fac[r - l] % mod, l = r)            while (r < val[i].size() && val[i][r] == val[i][l])                r ++;        cur = 1LL * cur * fac[n - val[i].size()] % mod;        ans = (ans + cur) % mod;    }    while (cnt)        ans = 1LL * ans * cnt % mod, cnt --;    return printf("%d\n", ans), 0;}