[whu1568]dp优化

http://acm.whu.edu.cn/land/problem/detail?problem_id=1568

思路：先将所有数分解，得到2,3,5,7的个数，转化为用这些2,3,5,7"构成"的不同序列的个数。一般思路，令dp[a][b][c][d]表示2有a个，3有b个，5有c个，7有d个时的答案，那么有如下转移方程:dp[a][b][c][d] = sigma（i:2->9）（a', b', c', d'）,a'为a减去i包含2的因子个数的结果,b',c',d'同理。由于空间消耗太大，必须另外考虑方法。注意到5和7是不可能组成其它的数的，可以单独处理，于是可以先用2和3构造，但需要把长度加进去作为状态的一部分，最后再把5和7插到构造成的序列里面。

 //#pragma comment(linker, "/STACK:10240000,10240000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << (a) << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

 typedef double db;
 typedef long long LL;
 typedef pair<int, int> pii;
 typedef multiset<int> msi;
 typedef set<int> si;
 typedef vector<int> vi;
 typedef map<int, int> mii;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 1e6 + ;
 const int md = 1e9 + ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }

 template<int mod>
 struct ModInt {
     const static int MD = mod;
     int x;
     ModInt(int x = ): x(x) {}
     int get() { return x; }

     ModInt operator + (const ModInt &that) const { int x0 = x + that.x; return ModInt(x0 < MD? x0 : x0 - MD); }
     ModInt operator - (const ModInt &that) const { int x0 = x - that.x; return ModInt(x0 < MD? x0 + MD : x0); }
     ModInt operator * (const ModInt &that) const { return ModInt((long long)x * that.x % MD); }
     ModInt operator / (const ModInt &that) const { return *this * that.inverse(); }

     ModInt operator += (const ModInt &that) { x += that.x; if (x >= MD) x -= MD; }
     ModInt operator -= (const ModInt &that) { x -= that.x; if (x < ) x += MD; }
     ModInt operator *= (const ModInt &that) { x = (long long)x * that.x % MD; }
     ModInt operator /= (const ModInt &that) { *this = *this / that; }

     ModInt inverse() const {
         int a = x, b = MD, u = , v = ;
         while(b) {
             int t = a / b;
             a -= t * b; std::swap(a, b);
             u -= t * v; std::swap(u, v);
         }
         if(u < ) u += MD;
         return u;
     }

 };
 typedef ModInt<> mint;

 const int c[][] = {{, }, {, }, {, }, {, }, {, }, {, }, {, }, {, }, {, }, {, }};
 const int d[] = {, };
 int cnt[];
 mint dp[][][];
 bool vis[][][];
 mint fact[], fact_inv[];

 mint dfs(int a, int b, int p) {
     if (vis[a][b][p]) return dp[a][b][p];
     if (p == ) return ;
     vis[a][b][p] = true;
     dp[a][b][p] = ;
     for (int i = ; i < ; i++) {
         if (i ==  || i == ) continue;
         int aa = c[i][], bb = c[i][];
         if (aa <= a && bb <= b) dp[a][b][p] += dfs(a - aa, b - bb, p - );
     }
     return dp[a][b][p];
 }

 void Init() {
     fact[] = ;
     fact_inv[] = ;
     for (int i = ; i <= ; i++) {
         fact[i] = fact[i - ] * i;
         fact_inv[i] = fact[i].inverse();
     }
 }

 int n;
 char s[maxn];

 int main() {
     //freopen("in.txt", "r", stdin);
     dp[][][] = ;
     vis[][][] = true;
     Init();
     while (cin >> n) {
         scanf("%s", s);
         mem0(cnt);
         rep_up0(i, n) {
             if (s[i] == '' || s[i] == '') {
                 cnt[s[i] - '']++;
                 continue;
             }
             rep_up0(j, ) {
                 cnt[d[j]] += c[s[i] - ''][j];
             }
         }
         mint ans = ;
         int c23 = cnt[] + cnt[];
         rep_up1(i, c23) ans += dfs(cnt[], cnt[], i) * fact[i + cnt[] + cnt[]] * fact_inv[cnt[]] * fact_inv[cnt[]] * fact_inv[i];
         printf("%d\n", ans.get());
     }
     return ;
 }