catalan数,c++实现入栈出栈顺序图

#include <iostream>
#include<algorithm>
#include<stack>
#include<vector>
using namespace std;
bool nixu(vector<int>cp) {
    
    for (size_t i = 2; i < cp.size(); ++i) {
        int max = cp[i - 1];
            size_t j = i;
        while (j < cp.size()) {
            if (cp[j++] > max)return false;
        }
    }
    return true;

}
auto delete_zero(vector<int>cp) {
    for (size_t i = 0; i < cp.size(); ++i)
    {
        if (cp[i] == 0)
        {
            cp.erase(cp.begin() + i);
            i--;
        }
        /*for (auto j : cp)
        cout << j << ends;
    cout << endl;*/

    }return cp;
}
bool set(vector<int>cp) {
    int first = *(cp.begin());
    for (size_t i = 1; i < cp.size(); ++i)
        if (cp[i] > first)cp[i] = 0;
    cp=delete_zero(cp);
    /*for (auto j : cp)
        cout << j << ends;
    cout << endl;*/
    return nixu(cp);
}

int main()
{
    int kibou=0;
    vector<int>num{ 1,2,3,4,5,6,7 };
    for (auto i : num)
        cout << i << ends;
    cout << endl;
    s: while (next_permutation(num.begin(), num.begin() + num.size())) {
        vector<int>newnum = num;
        for (auto i = newnum.size(); i > 2; --i) {
            if (!set(newnum))
                goto s;
            else
                newnum.erase(newnum.begin());
        }
        kibou++;
        for (auto i : num)
            cout << i << ends;
        cout << endl;
    }

    cout <<"出栈顺序数"<<++kibou << endl;
    

}

令h(0)=1,h(1)=1，卡塔兰数满足递归式：

h(n)= h(0)*h(n-1) + h(1)*h(n-2) + ... + h(n-1)h(0) (其中n>=2),这是n阶递推关系;

还可以化简为1阶递推关系: 如h(n)=(4n-2)/(n+1)*h(n-1)(n>1) h(0)=1

该递推关系的解为：h(n)=C(2n,n)/(n+1)=P(2n,n)/(n+1)!=(2n)!/(n!*(n+1)!) (n=1,2,3,...)

卡塔兰数列的前几项为(sequence A 0 0 0 1 0 8 in OEIS) [注： n = 0, 1, 2, 3, … n]

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, …