uva 10303 How Many Trees?
2016-01-22 13:42
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原题:
A binary search tree is a binary tree with root k such that any node v reachable from its left has
label(v) < label(k) and any node w reachable from its right has label(w) > label(k). It is a search
structure which can find a node with label x in O(nlogn) average time, where n is the size of the tree
(number of vertices).
Given a number n, can you tell how many different binary search trees may be constructed with a
set of numbers of size n such that each element of the set will be associated to the label of exactly one
node in a binary search tree?
Input
The input will contain a number 1 ≤ i ≤ 1000 per line representing the number of elements of the set.
Output
You have to print a line in the output for each entry with the answer to the previous question.
Sample Input
1
2
3
Sample Output
1
2
5
大意:
问你一个有n个节点的二叉查找树有多少种画法?
解答:
要是看过catalan数的话,直接就能判断出来。如果不太熟悉可以这样考虑,有n个有序的值x1,x2,x3….xn
现在以xk为根的话,xk左侧应该有k-1个节点,右侧有n-k个节点,设xk左侧节点的排列方法有f(k-1)种,右侧有f(n-k)种 最后以xk为根的二叉查找树就应该有f(k-1)*f(n-k)种方法,最后把所有xk求和就是catalan数。
大数模板:
转自/article/1422857.html
A binary search tree is a binary tree with root k such that any node v reachable from its left has
label(v) < label(k) and any node w reachable from its right has label(w) > label(k). It is a search
structure which can find a node with label x in O(nlogn) average time, where n is the size of the tree
(number of vertices).
Given a number n, can you tell how many different binary search trees may be constructed with a
set of numbers of size n such that each element of the set will be associated to the label of exactly one
node in a binary search tree?
Input
The input will contain a number 1 ≤ i ≤ 1000 per line representing the number of elements of the set.
Output
You have to print a line in the output for each entry with the answer to the previous question.
Sample Input
1
2
3
Sample Output
1
2
5
大意:
问你一个有n个节点的二叉查找树有多少种画法?
[code]#include<iostream> #include<algorithm> #include<map> #include<string> #include<cstring> #include<sstream> #include<cstdio> #include<vector> #include<cmath> #include<stack> #include<queue> #include<iomanip> #include<set> #include<fstream> using namespace std; #define MAXN 9999 #define MAXSIZE 10 #define DLEN 4 class BigNum { private: int a[500]; //可以控制大数的位数 int len; //大数长度 public: BigNum(){ len = 1;memset(a,0,sizeof(a)); } //构造函数 BigNum(const int); //将一个int类型的变量转化为大数 BigNum(const char*); //将一个字符串类型的变量转化为大数 BigNum(const BigNum &); //拷贝构造函数 BigNum &operator=(const BigNum &); //重载赋值运算符,大数之间进行赋值运算 friend istream& operator>>(istream&, BigNum&); //重载输入运算符 friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符 BigNum operator*(const BigNum &) const; //重载乘法运算符,两个大数之间的相乘运算 BigNum operator/(const int &) const; //重载除法运算符,大数对一个整数进行相除运算 void print(); //输出大数 }; BigNum::BigNum(const int b) //将一个int类型的变量转化为大数 { int c,d = b; len = 0; memset(a,0,sizeof(a)); while(d > MAXN) { c = d - (d / (MAXN + 1)) * (MAXN + 1); d = d / (MAXN + 1); a[len++] = c; } a[len++] = d; } BigNum::BigNum(const char*s) //将一个字符串类型的变量转化为大数 { int t,k,index,l,i; memset(a,0,sizeof(a)); l=strlen(s); len=l/DLEN; if(l%DLEN) len++; index=0; for(i=l-1;i>=0;i-=DLEN) { t=0; k=i-DLEN+1; if(k<0) k=0; for(int j=k;j<=i;j++) t=t*10+s[j]-'0'; a[index++]=t; } } BigNum::BigNum(const BigNum & T) : len(T.len) //拷贝构造函数 { int i; memset(a,0,sizeof(a)); for(i = 0 ; i < len ; i++) a[i] = T.a[i]; } BigNum & BigNum::operator=(const BigNum & n) //重载赋值运算符,大数之间进行赋值运算 { int i; len = n.len; memset(a,0,sizeof(a)); for(i = 0 ; i < len ; i++) a[i] = n.a[i]; return *this; } istream& operator>>(istream & in, BigNum & b) //重载输入运算符 { char ch[MAXSIZE*4]; int i = -1; in>>ch; int l=strlen(ch); int count=0,sum=0; for(i=l-1;i>=0;) { sum = 0; int t=1; for(int j=0;j<4&&i>=0;j++,i--,t*=10) { sum+=(ch[i]-'0')*t; } b.a[count]=sum; count++; } b.len =count++; return in; } ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符 { int i; cout << b.a[b.len - 1]; for(i = b.len - 2 ; i >= 0 ; i--) { cout.width(DLEN); cout.fill('0'); cout << b.a[i]; } return out; } BigNum BigNum::operator*(const BigNum & T) const //两个大数之间的相乘运算 { BigNum ret; int i,j,up; int temp,temp1; for(i = 0 ; i < len ; i++) { up = 0; for(j = 0 ; j < T.len ; j++) { temp = a[i] * T.a[j] + ret.a[i + j] + up; if(temp > MAXN) { temp1 = temp - temp / (MAXN + 1) * (MAXN + 1); up = temp / (MAXN + 1); ret.a[i + j] = temp1; } else { up = 0; ret.a[i + j] = temp; } } if(up != 0) ret.a[i + j] = up; } ret.len = i + j; while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret; } BigNum BigNum::operator/(const int & b) const //大数对一个整数进行相除运算 { BigNum ret; int i,down = 0; for(i = len - 1 ; i >= 0 ; i--) { ret.a[i] = (a[i] + down * (MAXN + 1)) / b; down = a[i] + down * (MAXN + 1) - ret.a[i] * b; } ret.len = len; while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret; } BigNum ans[1001]; int main() { ios::sync_with_stdio(false); int n; ans[1]=1; for(int i=2;i<=1000;i++) { ans[i]=ans[i-1]*(4*i-2)/(i+1); } while(cin>>n) { cout<<ans <<endl; } // input.close(); // output.close(); return 0; }
解答:
要是看过catalan数的话,直接就能判断出来。如果不太熟悉可以这样考虑,有n个有序的值x1,x2,x3….xn
现在以xk为根的话,xk左侧应该有k-1个节点,右侧有n-k个节点,设xk左侧节点的排列方法有f(k-1)种,右侧有f(n-k)种 最后以xk为根的二叉查找树就应该有f(k-1)*f(n-k)种方法,最后把所有xk求和就是catalan数。
大数模板:
[code]#include<iostream> #include<string> #include<iomanip> #include<algorithm> using namespace std; #define MAXN 9999 #define MAXSIZE 10 #define DLEN 4 class BigNum { private: int a[500]; //可以控制大数的位数 int len; //大数长度 public: BigNum(){ len = 1;memset(a,0,sizeof(a)); } //构造函数 BigNum(const int); //将一个int类型的变量转化为大数 BigNum(const char*); //将一个字符串类型的变量转化为大数 BigNum(const BigNum &); //拷贝构造函数 BigNum &operator=(const BigNum &); //重载赋值运算符,大数之间进行赋值运算 friend istream& operator>>(istream&, BigNum&); //重载输入运算符 friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符 BigNum operator+(const BigNum &) const; //重载加法运算符,两个大数之间的相加运算 BigNum operator-(const BigNum &) const; //重载减法运算符,两个大数之间的相减运算 BigNum operator*(const BigNum &) const; //重载乘法运算符,两个大数之间的相乘运算 BigNum operator/(const int &) const; //重载除法运算符,大数对一个整数进行相除运算 BigNum operator^(const int &) const; //大数的n次方运算 int operator%(const int &) const; //大数对一个int类型的变量进行取模运算 bool operator>(const BigNum & T)const; //大数和另一个大数的大小比较 bool operator>(const int & t)const; //大数和一个int类型的变量的大小比较 void print(); //输出大数 }; BigNum::BigNum(const int b) //将一个int类型的变量转化为大数 { int c,d = b; len = 0; memset(a,0,sizeof(a)); while(d > MAXN) { c = d - (d / (MAXN + 1)) * (MAXN + 1); d = d / (MAXN + 1); a[len++] = c; } a[len++] = d; } BigNum::BigNum(const char*s) //将一个字符串类型的变量转化为大数 { int t,k,index,l,i; memset(a,0,sizeof(a)); l=strlen(s); len=l/DLEN; if(l%DLEN) len++; index=0; for(i=l-1;i>=0;i-=DLEN) { t=0; k=i-DLEN+1; if(k<0) k=0; for(int j=k;j<=i;j++) t=t*10+s[j]-'0'; a[index++]=t; } } BigNum::BigNum(const BigNum & T) : len(T.len) //拷贝构造函数 { int i; memset(a,0,sizeof(a)); for(i = 0 ; i < len ; i++) a[i] = T.a[i]; } BigNum & BigNum::operator=(const BigNum & n) //重载赋值运算符,大数之间进行赋值运算 { int i; len = n.len; memset(a,0,sizeof(a)); for(i = 0 ; i < len ; i++) a[i] = n.a[i]; return *this; } istream& operator>>(istream & in, BigNum & b) //重载输入运算符 { char ch[MAXSIZE*4]; int i = -1; in>>ch; int l=strlen(ch); int count=0,sum=0; for(i=l-1;i>=0;) { sum = 0; int t=1; for(int j=0;j<4&&i>=0;j++,i--,t*=10) { sum+=(ch[i]-'0')*t; } b.a[count]=sum; count++; } b.len =count++; return in; } ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符 { int i; cout << b.a[b.len - 1]; for(i = b.len - 2 ; i >= 0 ; i--) { cout.width(DLEN); cout.fill('0'); cout << b.a[i]; } return out; } BigNum BigNum::operator+(const BigNum & T) const //两个大数之间的相加运算 { BigNum t(*this); int i,big; //位数 big = T.len > len ? T.len : len; for(i = 0 ; i < big ; i++) { t.a[i] +=T.a[i]; if(t.a[i] > MAXN) { t.a[i + 1]++; t.a[i] -=MAXN+1; } } if(t.a[big] != 0) t.len = big + 1; else t.len = big; return t; } BigNum BigNum::operator-(const BigNum & T) const //两个大数之间的相减运算 { int i,j,big; bool flag; BigNum t1,t2; if(*this>T) { t1=*this; t2=T; flag=0; } else { t1=T; t2=*this; flag=1; } big=t1.len; for(i = 0 ; i < big ; i++) { if(t1.a[i] < t2.a[i]) { j = i + 1; while(t1.a[j] == 0) j++; t1.a[j--]--; while(j > i) t1.a[j--] += MAXN; t1.a[i] += MAXN + 1 - t2.a[i]; } else t1.a[i] -= t2.a[i]; } t1.len = big; while(t1.a[len - 1] == 0 && t1.len > 1) { t1.len--; big--; } if(flag) t1.a[big-1]=0-t1.a[big-1]; return t1; } BigNum BigNum::operator*(const BigNum & T) const //两个大数之间的相乘运算 { BigNum ret; int i,j,up; int temp,temp1; for(i = 0 ; i < len ; i++) { up = 0; for(j = 0 ; j < T.len ; j++) { temp = a[i] * T.a[j] + ret.a[i + j] + up; if(temp > MAXN) { temp1 = temp - temp / (MAXN + 1) * (MAXN + 1); up = temp / (MAXN + 1); ret.a[i + j] = temp1; } else { up = 0; ret.a[i + j] = temp; } } if(up != 0) ret.a[i + j] = up; } ret.len = i + j; while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret; } BigNum BigNum::operator/(const int & b) const //大数对一个整数进行相除运算 { BigNum ret; int i,down = 0; for(i = len - 1 ; i >= 0 ; i--) { ret.a[i] = (a[i] + down * (MAXN + 1)) / b; down = a[i] + down * (MAXN + 1) - ret.a[i] * b; } ret.len = len; while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret; } int BigNum::operator %(const int & b) const //大数对一个int类型的变量进行取模运算 { int i,d=0; for (i = len-1; i>=0; i--) { d = ((d * (MAXN+1))% b + a[i])% b; } return d; } BigNum BigNum::operator^(const int & n) const //大数的n次方运算 { BigNum t,ret(1); int i; if(n<0) exit(-1); if(n==0) return 1; if(n==1) return *this; int m=n; while(m>1) { t=*this; for( i=1;i<<1<=m;i<<=1) { t=t*t; } m-=i; ret=ret*t; if(m==1) ret=ret*(*this); } return ret; } bool BigNum::operator>(const BigNum & T) const //大数和另一个大数的大小比较 { int ln; if(len > T.len) return true; else if(len == T.len) { ln = len - 1; while(a[ln] == T.a[ln] && ln >= 0) ln--; if(ln >= 0 && a[ln] > T.a[ln]) return true; else return false; } else return false; } bool BigNum::operator >(const int & t) const //大数和一个int类型的变量的大小比较 { BigNum b(t); return *this>b; } void BigNum::print() //输出大数 { int i; cout << a[len - 1]; for(i = len - 2 ; i >= 0 ; i--) { cout.width(DLEN); cout.fill('0'); cout << a[i]; } cout << endl; } int main(void) { int i,n; BigNum x[101]; //定义大数的对象数组 x[0]=1; for(i=1;i<101;i++) x[i]=x[i-1]*(4*i-2)/(i+1); while(scanf("%d",&n)==1 && n!=-1) { x .print(); } }
转自/article/1422857.html
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