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hdu 1130How Many Trees?(卡特兰数）

时间：2018-09-04 00:07:27      阅读：162      评论：0      收藏：0      [点我收藏+]

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卡特兰数又称卡塔兰数，英文名Catalan number，是组合数学中一个常出现在各种计数问题中出现的数列。

以比利时的数学家欧仁·查理·卡塔兰 (1814–1894)的名字来命名，其前几项为（从第零项开始） : 1, 1, 2, 5,

14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790,

477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324,

4861946401452, ...

卡特兰数有几条递推式：

    h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2)

   h(n)=h(n-1)*(4*n-2)/(n+1);

    h(n)=C(2n,n)/(n+1)

    h(n)=c(2n,n)-c(2n,n-1)

 

题目传送门：

题意：找规律

代码：

 

技术分享图片 
#include<iostream>
#include<vector>
#include<deque>
#include<string.h>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<string>
using namespace std;

class DividedByZeroException {};

class BigInteger {
private:
      vector<char> digits;
      bool sign;          //  true for positive, false for negitive
      void trim();        //  remove zeros in tail, but if the value is 0, keep only one:)
public:
      BigInteger(int);    // construct with a int integer
      BigInteger(string&) ;
      BigInteger();
      BigInteger(const BigInteger&);
      BigInteger operator=(const BigInteger& op2);

      BigInteger    abs() const;
      BigInteger    pow(int a);

      //binary operators

      friend BigInteger operator+=(BigInteger&, const BigInteger&);
      friend BigInteger operator-=(BigInteger&, const BigInteger&);
      friend BigInteger operator*=(BigInteger&, const BigInteger&);
      friend BigInteger operator/=(BigInteger&, const BigInteger&) throw(DividedByZeroException);
      friend BigInteger operator%=(BigInteger&, const BigInteger&) throw(DividedByZeroException);

      friend BigInteger operator+(const BigInteger&, const BigInteger&);
      friend BigInteger operator-(const BigInteger&, const BigInteger&);
      friend BigInteger operator*(const BigInteger&, const BigInteger&);
      friend BigInteger operator/(const BigInteger&, const BigInteger&) throw(DividedByZeroException);
      friend BigInteger operator%(const BigInteger&, const BigInteger&) throw(DividedByZeroException);


      //uniary operators
      friend BigInteger operator-(const BigInteger&);   //negative

      friend BigInteger operator++(BigInteger&);        //++v
      friend BigInteger operator++(BigInteger&, int);   //v++
      friend BigInteger operator--(BigInteger&);        //--v
      friend BigInteger operator--(BigInteger&, int);   //v--

      friend bool operator>(const BigInteger&, const BigInteger&);
      friend bool operator<(const BigInteger&, const BigInteger&);
      friend bool operator==(const BigInteger&, const BigInteger&);
      friend bool operator!=(const BigInteger&, const BigInteger&);
      friend bool operator>=(const BigInteger&, const BigInteger&);
      friend bool operator<=(const BigInteger&, const BigInteger&);

      friend ostream& operator<<(ostream&, const BigInteger&);   //print the BigInteger
      friend istream& operator>>(istream&, BigInteger&);         // input the BigInteger

public:
      static const BigInteger ZERO;
      static const BigInteger ONE;
      static const BigInteger TEN;
};
const BigInteger BigInteger::ZERO = BigInteger(0);
const BigInteger BigInteger::ONE = BigInteger(1);
const BigInteger BigInteger::TEN = BigInteger(10);


BigInteger::BigInteger() {
      sign = true;
}


BigInteger::BigInteger(int val) { // construct with a int integer
      if (val >= 0) {
            sign = true;
      }

      else {
            sign = false;
            val *= (-1);
      }

      do {
            digits.push_back((char)(val % 10));
            val /= 10;
      } while (val != 0);
}


BigInteger::BigInteger(string& def) {
      sign = true;

      for (string::reverse_iterator iter = def.rbegin() ; iter < def.rend();  iter++) {
            char ch = (*iter);

            if (iter == def.rend() - 1) {
                  if (ch == ‘+‘) {
                        break;
                  }

                  if (ch == ‘-‘) {
                        sign = false;
                        break;
                  }
            }

            digits.push_back((char)((*iter) - ‘0‘));
      }

      trim();
}

void BigInteger::trim() {
      vector<char>::reverse_iterator iter = digits.rbegin();

      while (!digits.empty() && (*iter) == 0) {
            digits.pop_back();
            iter = digits.rbegin();
      }

      if (digits.size() == 0) {
            sign = true;
            digits.push_back(0);
      }
}


BigInteger::BigInteger(const BigInteger& op2) {
      sign = op2.sign;
      digits = op2.digits;
}


BigInteger BigInteger::operator=(const BigInteger& op2) {
      digits = op2.digits;
      sign = op2.sign;
      return (*this);
}


BigInteger BigInteger::abs() const {
      if (sign) {
            return *this;
      }

      else {
            return -(*this);
      }
}

BigInteger BigInteger::pow(int a) {
      BigInteger res(1);

      for (int i = 0; i < a; i++) {
            res *= (*this);
      }

      return res;
}

//binary operators
BigInteger operator+=(BigInteger& op1, const BigInteger& op2) {
      if (op1.sign == op2.sign) {     //只处理相同的符号的情况，异号的情况给-处理
            vector<char>::iterator iter1;
            vector<char>::const_iterator iter2;
            iter1 = op1.digits.begin();
            iter2 = op2.digits.begin();
            char to_add = 0;        //进位

            while (iter1 != op1.digits.end() && iter2 != op2.digits.end()) {
                  (*iter1) = (*iter1) + (*iter2) + to_add;
                  to_add = ((*iter1) > 9);    // 大于9进一位
                  (*iter1) = (*iter1) % 10;
                  iter1++;
                  iter2++;
            }

            while (iter1 != op1.digits.end()) {    //
                  (*iter1) = (*iter1) + to_add;
                  to_add = ((*iter1) > 9);
                  (*iter1) %= 10;
                  iter1++;
            }

            while (iter2 != op2.digits.end()) {
                  char val = (*iter2) + to_add;
                  to_add = (val > 9) ;
                  val %= 10;
                  op1.digits.push_back(val);
                  iter2++;
            }

            if (to_add != 0) {
                  op1.digits.push_back(to_add);
            }

            return op1;
      }

      else {
            if (op1.sign) {
                  return op1 -= (-op2);
            }

            else {
                  return op1 = op2 - (-op1);
            }
      }

}

BigInteger operator-=(BigInteger& op1, const BigInteger& op2) {
      if (op1.sign == op2.sign) {     //只处理相同的符号的情况，异号的情况给+处理
            if (op1.sign) {
                  if (op1 < op2) { // 2 - 3
                        return  op1 = -(op2 - op1);
                  }
            }

            else {
                  if (-op1 > -op2) { // (-3)-(-2) = -(3 - 2)
                        return op1 = -((-op1) - (-op2));
                  }

                  else {           // (-2)-(-3) = 3 - 2
                        return op1 = (-op2) - (-op1);
                  }
            }

            vector<char>::iterator iter1;
            vector<char>::const_iterator iter2;
            iter1 = op1.digits.begin();
            iter2 = op2.digits.begin();

            char to_substract = 0;  //借位

            while (iter1 != op1.digits.end() && iter2 != op2.digits.end()) {
                  (*iter1) = (*iter1) - (*iter2) - to_substract;
                  to_substract = 0;

                  if ((*iter1) < 0) {
                        to_substract = 1;
                        (*iter1) += 10;
                  }

                  iter1++;
                  iter2++;
            }

            while (iter1 != op1.digits.end()) {
                  (*iter1) = (*iter1) - to_substract;
                  to_substract = 0;

                  if ((*iter1) < 0) {
                        to_substract = 1;
                        (*iter1) += 10;
                  }

                  else {
                        break;
                  }

                  iter1++;
            }

            op1.trim();
            return op1;
      }

      else {
            if (op1 > BigInteger::ZERO) {
                  return op1 += (-op2);
            }

            else {
                  return op1 = -(op2 + (-op1));
            }
      }
}
BigInteger operator*=(BigInteger& op1, const BigInteger& op2) {
      BigInteger result(0);

      if (op1 == BigInteger::ZERO || op2 == BigInteger::ZERO) {
            result = BigInteger::ZERO;
      }

      else {
            vector<char>::const_iterator iter2 = op2.digits.begin();

            while (iter2 != op2.digits.end()) {
                  if (*iter2 != 0) {
                        deque<char> temp(op1.digits.begin(), op1.digits.end());
                        char to_add = 0;
                        deque<char>::iterator iter1 = temp.begin();

                        while (iter1 != temp.end()) {
                              (*iter1) *= (*iter2);
                              (*iter1) += to_add;
                              to_add = (*iter1) / 10;
                              (*iter1) %= 10;
                              iter1++;
                        }

                        if (to_add != 0) {
                              temp.push_back(to_add);
                        }

                        int num_of_zeros = iter2 - op2.digits.begin();

                        while (num_of_zeros--) {
                              temp.push_front(0);
                        }

                        BigInteger temp2;
                        temp2.digits.insert(temp2.digits.end(), temp.begin(), temp.end());
                        temp2.trim();
                        result = result + temp2;
                  }

                  iter2++;
            }

            result.sign = ((op1.sign && op2.sign) || (!op1.sign && !op2.sign));
      }

      op1 = result;
      return op1;
}

BigInteger operator/=(BigInteger& op1, const BigInteger& op2) throw(DividedByZeroException) {
      if (op2 == BigInteger::ZERO) {
            throw DividedByZeroException();
      }

      BigInteger t1 = op1.abs(), t2 = op2.abs();

      if (t1 < t2) {
            op1 = BigInteger::ZERO;
            return op1;
      }

      //现在 t1 > t2 > 0
      //只需将 t1/t2的结果交给result就可以了
      deque<char> temp;
      vector<char>::reverse_iterator iter = t1.digits.rbegin();

      BigInteger temp2(0);

      while (iter != t1.digits.rend()) {
            temp2 = temp2 * BigInteger::TEN + BigInteger((int)(*iter));
            char s = 0;

            while (temp2 >= t2) {
                  temp2 = temp2 - t2;
                  s = s + 1;
            }

            temp.push_front(s);
            iter++;
      }

      op1.digits.clear();
      op1.digits.insert(op1.digits.end(), temp.begin(), temp.end());
      op1.trim();
      op1.sign = ((op1.sign && op2.sign) || (!op1.sign && !op2.sign));
      return op1;
}

BigInteger operator%=(BigInteger& op1, const BigInteger& op2) throw(DividedByZeroException) {
      return op1 -= ((op1 / op2) * op2);
}

BigInteger operator+(const BigInteger& op1, const BigInteger& op2) {
      BigInteger temp(op1);
      temp += op2;
      return temp;
}
BigInteger operator-(const BigInteger& op1, const BigInteger& op2) {
      BigInteger temp(op1);
      temp -= op2;
      return temp;
}

BigInteger operator*(const BigInteger& op1, const BigInteger& op2) {
      BigInteger temp(op1);
      temp *= op2;
      return temp;

}

BigInteger operator/(const BigInteger& op1, const BigInteger& op2) throw(DividedByZeroException) {
      BigInteger temp(op1);
      temp /= op2;
      return temp;
}

BigInteger operator%(const BigInteger& op1, const BigInteger& op2) throw(DividedByZeroException) {
      BigInteger temp(op1);
      temp %= op2;
      return temp;
}

//uniary operators
BigInteger operator-(const BigInteger& op) {  //negative
      BigInteger temp = BigInteger(op);
      temp.sign = !temp.sign;
      return temp;
}

BigInteger operator++(BigInteger& op) {   //++v
      op += BigInteger::ONE;
      return op;
}

BigInteger operator++(BigInteger& op, int x) { //v++
      BigInteger temp(op);
      ++op;
      return temp;
}

BigInteger operator--(BigInteger& op) {   //--v
      op -=  BigInteger::ONE;
      return op;
}

BigInteger operator--(BigInteger& op, int x) { //v--
      BigInteger temp(op);
      --op;
      return temp;
}

bool operator<(const BigInteger& op1, const BigInteger& op2) {
      if (op1.sign != op2.sign) {
            return !op1.sign;
      }

      else {
            if (op1.digits.size() != op2.digits.size())
                  return (op1.sign && op1.digits.size() < op2.digits.size())
                         || (!op1.sign && op1.digits.size() > op2.digits.size());

            vector<char>::const_reverse_iterator iter1, iter2;
            iter1 = op1.digits.rbegin();
            iter2 = op2.digits.rbegin();

            while (iter1 != op1.digits.rend()) {
                  if (op1.sign &&  *iter1 < *iter2) {
                        return true;
                  }

                  if (op1.sign &&  *iter1 > *iter2) {
                        return false;
                  }

                  if (!op1.sign &&  *iter1 > *iter2) {
                        return true;
                  }

                  if (!op1.sign &&  *iter1 < *iter2) {
                        return false;
                  }

                  iter1++;
                  iter2++;
            }

            return false;
      }
}
bool operator==(const BigInteger& op1, const BigInteger& op2) {
      if (op1.sign != op2.sign  || op1.digits.size() != op2.digits.size()) {
            return false;
      }

      vector<char>::const_iterator iter1, iter2;
      iter1 = op1.digits.begin();
      iter2 = op2.digits.begin();

      while (iter1 != op1.digits.end()) {
            if (*iter1 != *iter2) {
                  return false;
            }

            iter1++;
            iter2++;
      }

      return true;
}

bool operator!=(const BigInteger& op1, const BigInteger& op2) {
      return !(op1 == op2);
}

bool operator>=(const BigInteger& op1, const BigInteger& op2) {
      return (op1 > op2) || (op1 == op2);
}

bool operator<=(const BigInteger& op1, const BigInteger& op2) {
      return (op1 < op2) || (op1 == op2);
}

bool operator>(const BigInteger& op1, const BigInteger& op2) {
      return !(op1 <= op2);
}

ostream& operator<<(ostream& stream, const BigInteger& val) {  //print the BigInteger
      if (!val.sign) {
            stream << "-";
      }

      for (vector<char>::const_reverse_iterator iter = val.digits.rbegin(); iter != val.digits.rend() ; iter++) {
            stream << (char)((*iter) + ‘0‘);
      }

      return stream;
}

istream& operator>>(istream& stream, BigInteger& val) {   //Input the BigInteger
      string str;
      stream >> str;
      val = BigInteger(str);
      return stream;
}

#define maxn 105
BigInteger a[maxn];
int main() {
       int B;
       a[1]=1;
       for(int i=2;i<=maxn;i++)
       {
           a[i]=a[i-1]*(4*i-2)/(i+1);
       }
        while(cin >>B){
        cout<<a[B]<<endl;
        }
        /*cout << "A-B:" << A - B << endl;
        cout << "A+B:" << A + B << endl;
        cout << "A*B:" << A*B << endl;
        cout << "A/B:" << A / B << endl;
        cout << "A%B:" << A % B << endl;
        cout <<  A.pow(B-3) << endl;
        A++;
        cout << "A++:" << A << endl;
        A--;
        cout << "A--:" << A << endl;
        cout << "++B:" << ++B << endl;
        cout << "--B:" << --B << endl;
        cout << "C:" << C << endl;*/
}
卡特兰数+大数

 

卡特兰数的应用还有很多，可以借鉴这个博客：https://blog.csdn.net/zuzhiang/article/details/77966726( 里面有关于其在一些题目的应用 )

 

附上求卡特兰常数的模板：

 

技术分享图片 
const int C_maxn = 1e4 + 10;
LL CatalanNum[C_maxn];
LL inv[C_maxn];
inline void Catalan_Mod(int N, LL mod)
{
    inv[1] = 1;
    for(int i=2; i<=N+1; i++)///线性预处理 1 ~ N 关于 mod 的逆元
        inv[i] = (mod - mod / i) * inv[mod % i] % mod;

    CatalanNum[0] = CatalanNum[1] = 1;

    for(int i=2; i<=N; i++)
        CatalanNum[i] = CatalanNum[i-1] * (4 * i - 2) %mod * inv[i+1] %mod;
}
卡特兰常数模板

 

参考博客：http://www.cnblogs.com/Rubbishes/p/9468916.html

 

hdu 1130How Many Trees?(卡特兰数）

标签：logs   ons   vector   front   style   开始   ios   hide   bre   

原文地址：https://www.cnblogs.com/zhgyki/p/9581772.html
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