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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 123 |
Sample Output 125 |
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 12310 |
Sample Output 12516796 Hint The result will be very large, so you may not process it by 32-bit integers. |
#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("\n"); } 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!!