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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? |
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Input
The input will contain a number 1 <= i <= 100 per line representing the number of elements of the set.
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Output
You have to print a line in the output for each entry with the answer to the previous question.
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Sample Input
1 2 3 |
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Sample Output
1 2 5 |
View Code
1 #include <stdio.h> 2 int a[101][200] = {0}; 3 void catalan() 4 { 5 int carry, i, j, length, remainder; 6 a[1][1] = 1; 7 a[2][1] = 2; 8 a[3][1] = 5; 9 a[3][0] = 1; 10 for (i = 4; i <= 100; i++) 11 { 12 length = a[i - 1][0]; 13 carry = 0; 14 //handle multipling 15 for (j = 1; j <= length; j++) 16 { 17 carry += a[i - 1][j] * (4 * i - 2); 18 a[i][j] = carry % 10; 19 carry /= 10; 20 } 21 //determine the length 22 while (carry) 23 { 24 a[i][length++] = carry % 10; 25 carry /= 10; 26 } 27 remainder = 0; 28 //handle dividing 29 for (j = length; j >= 1; j--) 30 { 31 remainder = remainder * 10 + a[i][j]; 32 a[i][j] = remainder / (i + 1); 33 remainder %= (i + 1); 34 } 35 //determine the length 36 while (!a[i][length]) 37 { 38 length--; 39 } 40 a[i][0] = length; 41 } 42 } 43 int main() 44 { 45 int num, i; 46 catalan(); 47 while (scanf("%d", &num) != EOF) 48 { 49 if (num < 4) 50 printf("%d\n", a[num][1]); 51 else 52 { 53 for (i = a[num][0]; i >= 1; i--) 54 { 55 printf("%d", a[num][i]); 56 } 57 puts(""); 58 } 59 } 60 return 0; 61 }
Key points
firstly, you must realize that it's a calatan question. Then you can solve it.