Ultraviolet A 10312-Expression Bracketing (number theory + Catalan number)

Question link: 10312-Expression Bracketing question: there are n x numbers. brackets are required to determine the number of non-binary expressions. Idea: the calculation method of the Binary Expression is equal to the number of Catalan, so as long as the total number is calculated, the number of non-binary expressions is calculated by the total number minus the number of binary expressions. So what is the calculation method. Looking at the figure in the question, we can discuss n = 4 in the following situations: four 1, one 2, two 1, one 3, one 1, and the other 4, the numbers are 1, 3, 2, and 1. The answer is f (1) ^ 4 + 3 * f (2) * f (1) ^ 2 + f (3) * f (1) + f (4 ). One way is to break down n and calculate it, but obviously this is not feasible. n is at most 26, and there are too many cases. Then find the question and find that there is a formula. This formula is called the SuperCatalan number.

Then there is a recursive solution. If dp [n] [2] is set, n indicates that n subnodes are not allocated, and 2 indicates that 0 indicates that n-1 points are allocated at most, 1. allocate up to n vertices. This ensures that each subtree has at least two nodes. In this case, you can simply use the memory to search for the nodes.

Code:

Formula:

#include 
 
  #include 
  
   int n;long long Catalan[30], SuperCatalan[30];int main() {Catalan[1] = Catalan[2] = 1;for (int i = 3; i <= 26; i++) {Catalan[i] = Catalan[i - 1] * (4 * i - 6) / i; } SuperCatalan[1] = SuperCatalan[2] = 1; for (int i = 3; i <= 26; i++) {SuperCatalan[i] = (3 * (2 * i - 3) * SuperCatalan[i - 1] - (i - 3) * SuperCatalan[i - 2]) / i;}while (~scanf("%d", &n)) {printf("%lld\n", SuperCatalan[n] - Catalan[n]); }return 0;}
  
 

Recursive solution:

#include 
 
  #include 
  
   int n;long long Catalan[30], dp[30][2];long long dfs(int n, int flag) {long long &ans = dp[n][flag];if (~ans) return ans;if (n <= 1) return ans = 1;ans = 0;for (int i = 1; i < n + flag; i++)ans += dfs(i, 0) *dfs(n - i, 1);return ans;}int main() {Catalan[1] = Catalan[2] = 1;for (int i = 3; i <= 26; i++) {Catalan[i] = Catalan[i - 1] * (4 * i - 6) / i; }while (~scanf("%d", &n)) {memset(dp, -1, sizeof(dp));printf("%lld\n", dfs(n, 0) - Catalan[n]); }return 0;}