How Many Trees?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2761 Accepted Submission(s): 1619
Problem Description
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(n log n) 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?
with label x in O(n log n) 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 <= 100 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
/*
HDU 1130 How Many Trees?
据说是卡特兰数，那同1023
*/
#include<iostream>
using namespace std;
int a[101][101]={0};//a[i]表示的是第i个数
//因为是大数 [j]表示的是这个数的若干位
int main(){
int b[101],i,j,n,k,z;
a[1][0]=1;
b[1]=1;
k=1;//长度
for(i=2;i<101;i++)
{
for(j=0;j<k;j++)
a[i][j]=a[i-1][j]*(4*i-2);//乘法大数 每位乘以
z=0;//进位
for(j=0;j<k;j++)
{
a[i][j]+=z;
z=a[i][j]/10;
a[i][j]%=10;
}
while(z)//仍有进位
{
a[i][k++]=z%10;
z/=10;
}
//大数除法 模拟除法 从高到低
z=0;
for(j=k-1;j>=0;j--)
{
a[i][j]+=z*10;//上一位剩的
z=a[i][j]%(i+1);
a[i][j]/=(i+1);
}
while(!a[i][k-1])//去除前面的0
k--;
b[i]=k; //保存n的大数的长度
}
while(cin>>n)
{
for(i=b[n]-1;i>=0;i--)
cout<<a[n][i];
cout<<endl;
}
return 0;
}