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【题目链接】
【思路要点】
- 考虑用 引理计数。
- 不妨令 ,枚举 的整数拆分,在第二维上 即可。
- 具体来说,满足将 拆分为 的置换数为 ,一对环长为 的置换环对不动点个数产生的贡献为 ,枚举 的拆分后,可以通过简单 计算所需贡献。
- 令 ,时间复杂度 ,其中 表示 的整数拆分数。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 1005; const int P = 1e9 + 7; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int fac[MAXN], inv[MAXN], two[MAXN * MAXN], g[MAXN][MAXN]; int power(int x, int y) { if (y == 0) return 1; int tmp = power(x, y / 2); if (y % 2 == 0) return 1ll * tmp * tmp % P; else return 1ll * tmp * tmp % P * x % P; } int n, m, cnt, ans, a[MAXN], dp[MAXN][MAXN]; void update(int &x, int y) { x += y; if (x >= P) x -= P; } int getdp() { for (int i = 0; i <= m; i++) for (int j = 0; j <= m; j++) dp[i][j] = 0; dp[m][m] = 1; for (int i = m; i >= 0; i--) for (int j = m; j >= 1; j--) { if (dp[i][j] == 0) continue; int tmp = 0; for (int k = 1; k <= cnt; k++) tmp += g[j][a[k]]; int val = dp[i][j]; for (int k = 0; j * k <= i; k++) { update(dp[i - j * k][j - 1], 1ll * val * inv[k] % P); val = 1ll * val * inv[j] % P * fac[j - 1] % P * two[tmp] % P; } } return 1ll * dp[0][0] * fac[m] % P; } void check(int depth) { cnt = depth; int now = fac[n]; for (int i = 1; i <= cnt; i++) { now = 1ll * now * inv[a[i]] % P; now = 1ll * now * fac[a[i] - 1] % P; if (a[i] != a[i - 1]) { int j = i; while (j <= cnt && a[j] == a[i]) j++; now = 1ll * now * inv[j - i] % P; } } ans = (ans + 1ll * now * getdp()) % P; } void work(int lft, int lst, int depth) { if (lft == 0) { check(depth - 1); return; } if (lst == 0) return; work(lft, lst - 1, depth); if (lft >= lst) { a[depth++] = lst; work(lft - lst, lst, depth); } } int main() { read(n), read(m); fac[0] = inv[0] = 1; if (n > m) swap(n, m); for (int i = 1; i <= m; i++) { fac[i] = 1ll * fac[i - 1] * i % P; inv[i] = power(fac[i], P - 2); } two[0] = 1; for (int i = 1; i <= n * m; i++) two[i] = two[i - 1] * 2 % P; for (int i = 1; i <= m; i++) for (int j = 1; j <= m; j++) g[i][j] = __gcd(i, j); work(n, n, 1); writeln(1ll * ans * inv[n] % P * inv[m] % P); return 0; }