Codeforces 1109D Sasha and Interesting Fact from Graph Theory (看题解) 组合数学
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Sasha and Interesting Fact from Graph Theory
n 个 点形成 m 个有标号森林的方案数为 F(n, m) = m * n ^ {n - 1 - m}
然后就没啥难度了。。。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 1e6 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} int power(int a, int b) { int ans = 1; while(b) { if(b & 1) ans = 1LL * ans * a % mod; a = 1LL * a * a % mod; b >>= 1; } return ans; } int F[N], Finv[N], inv[N]; int C(int n, int m) { if(n < 0 || n < m) return 0; return 1LL * F[n] * Finv[m] % mod * Finv[n - m] % mod; } int n, m, a, b; int main() { inv[1] = F[0] = Finv[0] = 1; for(int i = 2; i < N; i++) inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod; for(int i = 1; i < N; i++) F[i] = 1LL * F[i - 1] * i % mod; for(int i = 1; i < N; i++) Finv[i] = 1LL * Finv[i - 1] * inv[i] % mod; scanf("%d%d%d%d", &n, &m, &a, &b); int ans = 0; for(int i = 2; i <= n; i++) { if(i < n) add(ans, 1LL * C(n - 2, i - 2) * F[i - 2] % mod * C(m - 1, i - 2) % mod * power(m, n - i) % mod * i % mod * power(n, n - i - 1) % mod); else add(ans, 1LL * F[i - 2] * C(m - 1, i - 2) % mod); } printf("%d\n", ans); return 0; } /* */
标签:const,Theory,int,题解,Finv,1LL,Interesting,mod,define 来源: https://www.cnblogs.com/CJLHY/p/10749474.html