Luogu P3182 [HAOI2016]放棋子

题目链接：​​传送门​​

题目说了
每行有一个障碍
两个障碍不在同一行也不在同一列
那障碍放哪里就没关系了
矩阵都不用输入
或者这样理解：
交换矩阵的某两行对答案是没有影响的
那么我们把所有障碍移到对角线处
那么剩下的棋子就是做一个错排

当然还有高精

#include <iostream>
#include <complex>
#include <cstring>
#include <cstdio>
#include <iomanip>

using namespace std;
namespace BigInteger {
  #define
  struct Big_integer {
    int d[maxn], len;
    void clean() {while(len > 1 and !d[len - 1]) len--;}
    Big_integer() {memset(d, 0, sizeof d); len = 1;}
    Big_integer(int num) {*this = num;}
    Big_integer(char* num) {*this = num;}
    Big_integer operator = (const char *num) {
      memset(d, 0, sizeof d);
      len = strlen(num);
      for (int i = 0; i < len; i++) d[i] = num[len - 1 - i] - '0';
      clean();
      return *this;
    }
    Big_integer operator = (int num) {
      char s[10005];
      sprintf(s, "%d", num);
      *this = s;
      return *this;
    }
    Big_integer operator + (const Big_integer &b) {
      Big_integer c = *this; int i;
      for (i = 0; i < b.len; i++) {
        c.d[i] += b.d[i];
        if (c.d[i] > 9)  c.d[i] %= 10, c.d[i + 1]++;
      }
      while (c.d[i] > 9) c.d[i++] %= 10, c.d[i]++;
      c.len = max(len, b.len);
      if (c.d[i] and c.len <= i) c.len = i + 1;
      return c;
    }
    Big_integer operator * (const Big_integer &b) const {
      int i, j; Big_integer c;
      c.len = len + b.len;
      for (j = 0; j < b.len; j++)
        for (i = 0; i < len; i++)
          c.d[i + j] += d[i] * b.d[j];
      for (i = 0; i < c.len - 1; i++) {
        c.d[i + 1] += c.d[i] / 10;
        c.d[i] %= 10;
      }
      c.clean();
      return c;
    }
    string str() const {
      char s[maxn];
      for (int i = 0; i < len; i++) s[len - 1 - i] = d[i] + '0';
      return s;
    }
  };
  istream& operator >> (istream &in, Big_integer &x) {
    string s;
    in >> s;
    x = s.c_str();
    return in;
  }
  ostream& operator << (ostream &out, const Big_integer &x) {
    out << x.str();
    return out;
  }
}
using namespace BigInteger;
Big_integer f[201]; int n;

int main(int argc, char const *argv[]) {
  cin >> n; f[1] = 0; f[2] = 1;
  for (int i = 3; i <= n; i++) f[i] = (f[i - 1] + f[i - 2]) * (i - 1);
  cout << f[n] << endl;
}