UVA 10312-expression bracketing (number theory +catalan)

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option=com_onlinejudge&itemid=8&page=show_problem&problem=1253 ">10312-expression Bracketing Test instructions: There are n x, which requires parentheses to infer the number of non-binary expressions. train of thought: Two the expression of the fork is equal to the number of Catalan, then only the total number is calculated, minus the number of binary expressions. The number of non-binary expressions is obtained.

So what is the calculation method? look at the graph in the topic, for the case of n = 4, can be divided into the following situations to discuss: four of 1. A 22 a 1, a 31 1. A 4. The corresponding number of cases is 1. 3. 2. 1. The answer is F (1) ^4 + 3 * F (2) * F (1) ^2 + f (3) * F (1) + F (4). One approach is to decompose N and then calculate. But obviously this is not feasible, n maximum is 26, the number of cases is too many. then find the puzzle, found that there is a formula, this is called Supercatalan number.

Then there are the solutions that are handed out. Set dp[n][2]. n indicates that there are also n sub-nodes that are not assigned. 2 means 0 is allocated a maximum of n-1 points, and 1 is allocated at most n points, which guarantees that the subtree has at least two nodes. This is the general situation, the direct use of memory search down can

Code:

Formula Solution:

#include <stdio.h> #include <string.h>int n;long long catalan[30], Supercatalan[30];int main () {catalan[1] = CATALAN[2] = 1;for (int i = 3; I <=; i++) {catalan[i] = catalan[i-1] * (4 * i-6)/I;} SUPERCATALAN[1] = supercatalan[2] = 1; for (int i = 3; I <=; 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 <stdio.h> #include <string.h>int n;long long catalan[30], Dp[30][2];long long dfs (int n, int. flag) {Lo  ng 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 <=; i++) {catalan[i] = catalan[i-1] * (4 * i-6)/I;} while (~SCANF ("%d", &n)) {memset (DP,-1, sizeof (DP));p rintf ("%lld\n", DFS (n, 0)-catalan[n]);} return 0;}

UVA 10312-expression bracketing (number theory +catalan)