题目大意：有$n$个点，问有多少种连成生成树的方案。
题解：根据$prufer$序列可得，$n$个点的生成树有$n^{n-2}$个，每种生成树有$(n-1)!$种生成方案，所以答案是$n^{n-2}(n-1)!$
卡点：无
C++ Code：
#include <cstdio>
const int mod = 9999991;
int n, sum;
inline int pw(int base, int p) {
static int res;
for (res = 1; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
return res;
}
int main() {
scanf("%d", &n);
sum = pw(n, n - 2);
for (int i = 2; i < n; ++i) sum = static_cast<long long> (sum) * i % mod;
printf("%d\n", sum);
return 0;
}
原文：https://www.cnblogs.com/Memory-of-winter/p/10326713.html