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摘要: long long f[25]; int main() { f[0] = 1; cin >> n; for (int i = 1; i <= n; i++) f[i] = f[i - 1] * (4 * i - 2) / (i + 1); //这里用的是常见公式2 cout << f[n] << e 阅读全文
posted @ 2020-02-19 14:39 MQFLLY 阅读(169) 评论(0) 推荐(0) 编辑