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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int mod = 1e9 + 7;
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const int N = 110;
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int f[N];
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int qmi(int a, int k, int p) {
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int res = 1;
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while (k) {
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if (k & 1) res = (LL)res * a % p;
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a = (LL)a * a % p;
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k >>= 1;
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}
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return res;
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}
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int C(int a, int b, int p) {
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if (a < b) return 0;
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int down = 1, up = 1;
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for (int i = a, j = 1; j <= b; i--, j++) {
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up = (LL)up * i % p;
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down = (LL)down * j % p;
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}
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return (LL)up * qmi(down, p - 2, p) % p;
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}
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//卡特兰数的前几项为：1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786,208012
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int main() {
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//目标：求出卡特兰数的前12位
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//公式法计算I
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//方法1:f(n)=C(n,2n)-C(n-1,2n)
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for (int i = 1; i <= 12; i++)
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cout << C(2 * i, i, mod) - C(2 * i, i - 1, mod) << " ";
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cout << endl;
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//公式法计算II
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//方法2：C(n,2n)/(n+1)
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for (int i = 1; i <= 12; i++)
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cout << C(2 * i, i, mod) / (i + 1) << " ";
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cout << endl;
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//递推法计算I
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//方法3：f(n)=(4n-2)/(n+1)*f(n-1)
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f[0] = 1;
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for (int i = 1; i <= 12; i++) {
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f[i] = (LL)f[i - 1] * (4 * i - 2) / (i + 1); //注意：这里要先乘再除，否则可能丢失精度
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cout << f[i] << " ";
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}
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cout << endl;
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//递推法计算II
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//方法4： f(n)=\sum_{i=0}^{n-1}f(i)*f(n-i+1)
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//要是每次都这么推，不是效率太差了吗？
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memset(f, 0, sizeof f);
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f[0] = 1;
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for (int i = 1; i <= 12; i++) {
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for (int j = 1; j <= i; j++)
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f[i] += f[j - 1] * f[i - j];
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cout << f[i] << " ";
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}
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return 0;
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} |
