[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。

  1 //#pragma comment(linker, "/STACK:10240000,10240000")
  2 
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <set>
 16 #include <bitset>
 17 #include <functional>
 18 #include <numeric>
 19 #include <stdexcept>
 20 #include <utility>
 21 
 22 using namespace std;
 23 
 24 #define mem0(a) memset(a, 0, sizeof(a))
 25 #define mem_1(a) memset(a, -1, sizeof(a))
 26 #define lson l, m, rt << 1
 27 #define rson m + 1, r, rt << 1 | 1
 28 #define define_m int m = (l + r) >> 1
 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 32 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 33 #define all(a) (a).begin(), (a).end()
 34 #define lowbit(x) ((x) & (-(x)))
 35 #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) {}
 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 38 #define pchr(a) putchar(a)
 39 #define pstr(a) printf("%s", a)
 40 #define sstr(a) scanf("%s", a)
 41 #define sint(a) scanf("%d", &a)
 42 #define sint2(a, b) scanf("%d%d", &a, &b)
 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 44 #define pint(a) printf("%d\n", a)
 45 #define test_print1(a) cout << "var1 = " << (a) << endl
 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 48 
 49 typedef double db;
 50 typedef long long LL;
 51 typedef pair<int, int> pii;
 52 typedef multiset<int> msi;
 53 typedef set<int> si;
 54 typedef vector<int> vi;
 55 typedef map<int, int> mii;
 56 
 57 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
 58 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
 59 const int maxn = 1e6 + 7;
 60 const int md = 1e9 + 7;
 61 const int inf = 1e9 + 7;
 62 const LL inf_L = 1e18 + 7;
 63 const double pi = acos(-1.0);
 64 const double eps = 1e-6;
 65 
 66 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 67 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 68 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 69 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 70 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 71 int make_id(int x, int y, int n) { return x * n + y; }
 72 
 73 template<int mod>
 74 struct ModInt {
 75     const static int MD = mod;
 76     int x;
 77     ModInt(int x = 0): x(x) {}
 78     int get() { return x; }
 79 
 80     ModInt operator + (const ModInt &that) const { int x0 = x + that.x; return ModInt(x0 < MD? x0 : x0 - MD); }
 81     ModInt operator - (const ModInt &that) const { int x0 = x - that.x; return ModInt(x0 < MD? x0 + MD : x0); }
 82     ModInt operator * (const ModInt &that) const { return ModInt((long long)x * that.x % MD); }
 83     ModInt operator / (const ModInt &that) const { return *this * that.inverse(); }
 84 
 85     ModInt operator += (const ModInt &that) { x += that.x; if (x >= MD) x -= MD; }
 86     ModInt operator -= (const ModInt &that) { x -= that.x; if (x < 0) x += MD; }
 87     ModInt operator *= (const ModInt &that) { x = (long long)x * that.x % MD; }
 88     ModInt operator /= (const ModInt &that) { *this = *this / that; }
 89 
 90     ModInt inverse() const {
 91         int a = x, b = MD, u = 1, v = 0;
 92         while(b) {
 93             int t = a / b;
 94             a -= t * b; std::swap(a, b);
 95             u -= t * v; std::swap(u, v);
 96         }
 97         if(u < 0) u += MD;
 98         return u;
 99     }
100 
101 };
102 typedef ModInt<1000000007> mint;
103 
104 
105 const int c[10][2] = {{0, 0}, {0, 0}, {1, 0}, {0, 1}, {2, 0}, {0, 0}, {1, 1}, {0, 0}, {3, 0}, {0, 2}};
106 const int d[2] = {2, 3};
107 int cnt[10];
108 mint dp[160][110][160];
109 bool vis[160][110][160];
110 mint fact[420], fact_inv[420];
111 
112 mint dfs(int a, int b, int p) {
113     if (vis[a][b][p]) return dp[a][b][p];
114     if (p == 0) return 0;
115     vis[a][b][p] = true;
116     dp[a][b][p] = 0;
117     for (int i = 2; i < 10; i++) {
118         if (i == 5 || i == 7) continue;
119         int aa = c[i][0], bb = c[i][1];
120         if (aa <= a && bb <= b) dp[a][b][p] += dfs(a - aa, b - bb, p - 1);
121     }
122     return dp[a][b][p];
123 }
124 
125 void Init() {
126     fact[0] = 1;
127     fact_inv[0] = 1;
128     for (int i = 1; i <= 410; i++) {
129         fact[i] = fact[i - 1] * i;
130         fact_inv[i] = fact[i].inverse();
131     }
132 }
133 
134 int n;
135 char s[maxn];
136 
137 int main() {
138     //freopen("in.txt", "r", stdin);
139     dp[0][0][0] = 1;
140     vis[0][0][0] = true;
141     Init();
142     while (cin >> n) {
143         scanf("%s", s);
144         mem0(cnt);
145         rep_up0(i, n) {
146             if (s[i] == '5' || s[i] == '7') {
147                 cnt[s[i] - '0']++;
148                 continue;
149             }
150             rep_up0(j, 2) {
151                 cnt[d[j]] += c[s[i] - '0'][j];
152             }
153         }
154         mint ans = 0;
155         int c23 = cnt[2] + cnt[3];
156         rep_up1(i, c23) ans += dfs(cnt[2], cnt[3], i) * fact[i + cnt[5] + cnt[7]] * fact_inv[cnt[5]] * fact_inv[cnt[7]] * fact_inv[i];
157         printf("%d\n", ans.get());
158     }
159     return 0;
160 }
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posted @ 2015-04-22 20:14  jklongint  阅读(346)  评论(1编辑  收藏  举报