Uva10312-expression bracketing (Catalan + recursion)

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Question: Give a sequence with a length of N, which indicates that there are N x (nodes). You can add any parentheses and ask how many strings are in a non-Binary Expression.

Idea: subtract the number of binary expressions from the total number. The binary expression can be solved using the catalan number. For the total number, use the DP. DP [I] [0] indicates that it can be split into two Subtrees at the position I. DP [I] [1] indicates that there is a subtree.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int MAXN = 30;ll Catalan[MAXN], dp[MAXN][2];int n;void init() {    memset(Catalan, 0, sizeof(Catalan));    Catalan[1] = Catalan[2] = 1;    for (int i = 3; i < MAXN; i++)        Catalan[i] = Catalan[i - 1] * (4 * i - 6) / i; }ll dfs(int n, int m) {    ll& ans = dp[n][m];    if (ans != 0)        return ans;    if (n <= 1)        return 1;    ans = 0;    for (int i = 1; i < n + m; i++)        ans += dfs(i, 1) * dfs(n - i, 0);    return ans;}int main() {    init();    while (scanf("%d", &n) != EOF) {        memset(dp, 0, sizeof(dp));         printf("%lld\n", dfs(n, 0) - Catalan[n]);    }    return 0;}

Uva10312-expression bracketing (Catalan + recursion)