bzoj【Kirchhoff矩阵】

来源：互联网发布：他知而自不知的我属于编辑：程序博客网时间：2024/06/07 01:52
学习：2007年周冬的论文。
题意：中文题意。
题解：推结论推出f[i]=(f[i-1]*3-f[i-2]+2)，用高精度模拟。
代码：
/**************************************************************    Problem: 1002    User: Jstyle    Language: C++    Result: Accepted    Time:44 ms    Memory:1456 kb****************************************************************/ #include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <bitset>#include <vector>#include <queue>#include <stack>#include <cmath>#include <list>#include <set>#include <map>#define rep(i,a,b) for(int (i) = (a);(i) <= (b);++ (i))#define per(i,a,b) for(int (i) = (a);(i) >= (b);-- (i))#define mem(a,b) memset((a),(b),sizeof((a)))#define FIN freopen("in.txt","r",stdin)#define FOUT freopen("out.txt","w",stdout)#define IO ios_base::sync_with_stdio(0),cin.tie(0)#define mid ((l+r)>>1)#define ls (id<<1)#define rs ((id<<1)|1)#define N 100000+5#define INF 0x3f3f3f3f#define INFF 0x3f3f3f3f3f3f3ftypedef long long ll;const ll mod = 20071027;const ll eps = 1e-12;using namespace std; const int MAXN = 410;     struct bign  {      int len, s[MAXN];      bign ()      {          memset(s, 0, sizeof(s));          len = 1;      }      bign (int num) { *this = num; }      bign (const char *num) { *this = num; }      bign operator = (const int num)      {          char s[MAXN];          sprintf(s, "%d", num);          *this = s;          return *this;      }      bign operator = (const char *num)      {          for(int i = 0; num[i] == '0'; num++) ;  //去前导0          len = strlen(num);          for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';          return *this;      }      bign operator + (const bign &b) const //+      {          bign c;          c.len = 0;          for(int i = 0, g = 0; g || i < max(len, b.len); i++)          {              int x = g;              if(i < len) x += s[i];              if(i < b.len) x += b.s[i];              c.s[c.len++] = x % 10;              g = x / 10;          }          return c;      }      bign operator += (const bign &b)      {          *this = *this + b;          return *this;      }      void clean()      {          while(len > 1 && !s[len-1]) len--;      }      bign operator * (const bign &b) //*      {          bign c;          c.len = len + b.len;          for(int i = 0; i < len; i++)          {              for(int j = 0; j < b.len; j++)              {                  c.s[i+j] += s[i] * b.s[j];              }          }          for(int i = 0; i < c.len; i++)          {              c.s[i+1] += c.s[i]/10;              c.s[i] %= 10;          }          c.clean();          return c;      }      bign operator *= (const bign &b)      {          *this = *this * b;          return *this;      }      bign operator - (const bign &b)      {          bign c;          c.len = 0;          for(int i = 0, g = 0; i < len; i++)          {              int x = s[i] - g;              if(i < b.len) x -= b.s[i];              if(x >= 0) g = 0;              else             {                  g = 1;                  x += 10;              }              c.s[c.len++] = x;          }          c.clean();          return c;      }      bign operator -= (const bign &b)      {          *this = *this - b;          return *this;      }      bign operator / (const bign &b)      {          bign c, f = 0;          for(int i = len-1; i >= 0; i--)          {              f = f*10;              f.s[0] = s[i];              while(f >= b)              {                  f -= b;                  c.s[i]++;              }          }          c.len = len;          c.clean();          return c;      }      bign operator /= (const bign &b)      {          *this  = *this / b;          return *this;      }      bign operator % (const bign &b)      {          bign r = *this / b;          r = *this - r*b;          return r;      }      bign operator %= (const bign &b)      {          *this = *this % b;          return *this;      }      bool operator < (const bign &b)      {          if(len != b.len) return len < b.len;          for(int i = len-1; i >= 0; i--)          {              if(s[i] != b.s[i]) return s[i] < b.s[i];          }          return false;      }      bool operator > (const bign &b)      {          if(len != b.len) return len > b.len;          for(int i = len-1; i >= 0; i--)          {              if(s[i] != b.s[i]) return s[i] > b.s[i];          }          return false;      }      bool operator == (const bign &b)      {          return !(*this > b) && !(*this < b);      }      bool operator != (const bign &b)      {          return !(*this == b);      }      bool operator <= (const bign &b)      {          return *this < b || *this == b;      }      bool operator >= (const bign &b)      {          return *this > b || *this == b;      }      string str() const     {          string res = "";          for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;          return res;      }  };     istream& operator >> (istream &in, bign &x)  {      string s;      in >> s;      x = s.c_str();      return in;  }     ostream& operator << (ostream &out, const bign &x)  {      out << x.str();      return out;  }   int n;bign dp[105];void fuc(){    bign a = "3",b = "2";    dp[1] = "1";    dp[2] = "5";    rep(i, 3, 100)        dp[i] = a*dp[i-1]-dp[i-2]+b;}int main()  {    fuc();    while(cin >> n)        cout << dp[n] << endl;    return 0;  }
0 0