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发表于 2022-6-5 13:23:20
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我还真的写出来了,哈哈哈 :- #include <iostream>
- #include <cmath>
- namespace StirlingNumber {
- // 階乘
- unsigned long long f(int n) {
- if (not n) return 1;
- return n * f(n - 1);
- }
- // 组合公式
- unsigned long long C(int n, int k) {
- return (f(n)) / (f(k) * f(n - k));
- }
- // 第二类斯特林数
- unsigned long long S(int n, int m) {
- if (not n) return 1;
- else if (not m) return static_cast<unsigned long long> (pow(m, n));
- else if (m == 1 or n == m) return m;
- else if (m == 2) return static_cast<unsigned long long> (pow(2, n - 1)) - 1;
- else if (m == n - 1)return C(n, 2);
- else if (m == n - 2) return C(n, 3) + 3 * C(n, 4);
- else if (m == 3) return static_cast<unsigned long long> (.5 * pow(3, n - 1) - pow(2, n - 1));
- else if (m == n - 3) return C(n, 4) + 10 * C(n, 5) + 15 * C(n, 6);
- return S(n - 1, m - 1) + m * S(n - 1, m);
- }
- }
- using namespace StirlingNumber;
- using std::cout, std::endl;
- int main(void) {
- cout << S(15, 12) << endl;
- return 0;
- }
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