How Many Trees? |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 1105 Accepted Submission(s): 577 |
|
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? |
|
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 |
Train Problem II |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 730 Accepted Submission(s): 430 |
|
Problem Description
As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.
|
|
Input
The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.
|
|
Output
For each test case, you should output how many ways that all the trains can get out of the railway. |
|
Sample Input
1 2 3 10 |
|
Sample Output
1 2 5 16796 |
#include<iostream>
#include<cstdio>#include<cstring>
int h[1001][1001];
int Catlan()
{
memset(h,0,sizeof(h));
h[1][0]=1;
h[1][1]=1;
h[2][0]=1;
h[2][1]=2;
for(int i=3;i<=101;i++)
{
int len=h[i-1][0];
for(int j=1;j<=len;j++)
{
h[i][j]+=h[i-1][j]*(4*i-2);
if(h[i][j]>=10)
{
h[i][j+1]+=h[i][j]/10;
h[i][j]%=10;
}
}
len=h[i][len+1]==0?len:len+1;
while(h[i][len]>=10)
{
h[i][1+len]=h[i][len]/10;
h[i][len]%=10;
len++;
}
int yu=0;
for(int k=len;k>=1;k--)
{
int temp=(h[i][k]+yu*10)/(i+1);
yu=(h[i][k]+yu*10)%(i+1);
h[i][k]=temp;
}
while(!h[i][len])
len--;
h[i][0]=len;
}
}
int main()
{
int n;
Catlan();
while(~scanf("%d",&n))
{
for(int i=h[n][0];i>=1;i--)
printf("%d",h[n][i]);
printf(" ");
}
return 0;
}
/******两个典型的卡塔兰数模板题（模板到代码都可以一样），火车那个的话以0表示出栈，1表示进栈，就看得出来了，二叉树的话，假设N个节点，根节点必须要有一个，接下来，可以左边0个，右边n-1个或左边1个，右边N-1个
或，，，，左边N-1个，右边0个，也看得出来是卡塔兰
万圣节之夜脱单的同学都和女票出去了，还在刷题，，Orz
不过刷题的感觉还是蛮不错的，哈哈，喜欢这种运动+刷题+和朋友在一起+泡图书馆的感觉！！