[容斥原理][DP] SRM 498 Div1 Hard FoxJumping

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$Solution$

对不能走 $(+a_i,+a_i)$ 容斥，设 $dp_{p,q}$ 表示走了 $p$ 步不合法的，和为 $10q$ 的方案数。
同时DP出 $x,y$ 两个方向上走了 $p$ 步，长度为 $q$ 的方案数 $x_{p,q},y_{p,q}$ 。因为此时 $x,y$ 独立，最后计算的时候直接乘起来就好了。
那最后答案就是

\sum p = 0 R \sum q = 0 ⌊ m i n ( m , n ) 10 ⌋ (- 1) p (R p) x R - p, n - 10 p y R - p, m - 10 q d p p, q

$\sum_{p=0}^R\sum_{q=0}^{\lfloor{min(m,n)\over10}\rfloor}(-1)^p{R\choose p}x_{R-p,n-10p}y_{R-p,m-10q}dp_{p,q}$

x,y $x,y$ 的DP显然可以前缀和优化，~~刚开始暴力n3T掉以后还打算FFT~~。

// BEGIN CUT HERE

// END CUT HERE
#line 5 "FoxJumping.cpp"
#include <bits/stdc++.h>
using namespace std;

const int N = 888;
const int MOD = 10007;

int dp[N << 1][N];
int x[N << 1][N], y[N << 1][N];
int gx[N << 1][N], gy[N << 1][N];
int C[N << 1][N << 1];
int n, m, mn, dx, dy, R, cnt, ans;
int a[N];

class FoxJumping {

public:
  inline void Add(int &x, int a) {
    x = (x + a) % MOD;
  }
  int getCount(int Tx, int Ty, int Mx, int My, int _R, vector <int> bad) {
    n = Tx; m = Ty; dx = Mx; dy = My; R = _R; mn = min(m, n);
    bad.push_back(0); cnt = bad.size();
    for (int i = 0; i < cnt; i++) a[i + 1] = bad[i];
    sort(a + 1, a + cnt + 1);
    cnt = unique(a + 1, a + cnt + 1) - a - 1;

    ans = 0;
    memset(x, 0, sizeof x);
    memset(y, 0, sizeof y);
    memset(dp, 0, sizeof dp);

    for (int i = 0; i <= R; 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;
    }

    x[0][0] = 1;
    for (int i = 0; i <= R; i++)
      for (int j = 0; j <= n; j++) {
        gx[i][j + 1] = (gx[i][j] + x[i][j]) % MOD;
        Add(x[i + 1][j], (gx[i][j + 1] - gx[i][max(j + 1 - dx - 1, 0)] + MOD) % MOD);
      }
    y[0][0] = 1;
    for (int i = 0; i <= R; i++)
      for (int j = 0; j <= m; j++) {
        gy[i][j + 1] = (gy[i][j] + y[i][j]) % MOD;
        Add(y[i + 1][j], (gy[i][j + 1] - gy[i][max(j + 1 - dy - 1, 0)] + MOD) % MOD);
      }
    dp[0][0] = 1;
    for (int i = 0; i <= R; i++)
      for (int j = 0; j * 10 <= mn; j++)
        for (int k = 1; k <= cnt; k++)
          if (j * 10 + a[k] <= mn) Add(dp[i + 1][j + a[k] / 10], dp[i][j]);
          else break;

    for (int p = 0; p <= R; p++)
      for (int q = 0; 10 * q <= mn; q++) {
        int res = x[R - p][n - 10 * q] * y[R - p][m - 10 * q] % MOD * dp[p][q] % MOD * C[R][p] % MOD;
        if (p & 1) Add(ans, MOD - res);
        else Add(ans, res);
      }
    return ans;
  }

  // BEGIN CUT HERE
public:
  void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) test_case_5(); }
private:
  template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
  void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
  void test_case_0() { int Arg0 = 2; int Arg1 = 2; int Arg2 = 1; int Arg3 = 1; int Arg4 = 2; int Arr5[] = {}; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); int Arg6 = 1; verify_case(0, Arg6, getCount(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
  void test_case_1() { int Arg0 = 2; int Arg1 = 2; int Arg2 = 1; int Arg3 = 1; int Arg4 = 3; int Arr5[] = {}; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); int Arg6 = 6; verify_case(1, Arg6, getCount(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
  void test_case_2() { int Arg0 = 10; int Arg1 = 10; int Arg2 = 10; int Arg3 = 10; int Arg4 = 1; int Arr5[] = {}; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); int Arg6 = 1; verify_case(2, Arg6, getCount(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
  void test_case_3() { int Arg0 = 10; int Arg1 = 10; int Arg2 = 10; int Arg3 = 10; int Arg4 = 1; int Arr5[] = {10}; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); int Arg6 = 0; verify_case(3, Arg6, getCount(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
  void test_case_4() { int Arg0 = 11; int Arg1 = 11; int Arg2 = 11; int Arg3 = 11; int Arg4 = 2; int Arr5[] = {10}; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); int Arg6 = 140; verify_case(4, Arg6, getCount(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
  void test_case_5() { int Arg0 = 123; int Arg1 = 456; int Arg2 = 70; int Arg3 = 80; int Arg4 = 90; int Arr5[] = {30, 40, 20, 10, 50}; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); int Arg6 = 6723; verify_case(5, Arg6, getCount(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }

  // END CUT HERE


};

// BEGIN CUT HERE
int main(void) {
  FoxJumping ___test;
  ___test.run_test(-1);
  system("pause");
}
// END CUT HERE