版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/qq_39972971/article/details/88389881
【题目链接】
【思路要点】
- 记 表示数列的前 项的并有 位,枚举数列的第 位使得数值的并增加了 位,则将 加入 。
- 注意到转移与 无关,我们可以考虑将该 的第一维通过倍增优化。
- 考虑通过 得到 ,枚举数列的前 项的并的位数 ,有:
- 写成卷积的形式即为:
- 任意模数 优化即可。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 262144; const int MAXLOG = 15; 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(""); } namespace AnyModuloFFT { const int MAXN = 262144; const long double pi = acosl(-1); struct point {long double x, y; }; point operator + (point a, point b) {return (point) {a.x + b.x, a.y + b.y}; } point operator - (point a, point b) {return (point) {a.x - b.x, a.y - b.y}; } point operator * (point a, point b) {return (point) {a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x}; } point operator / (point a, long double x) {return (point) {a.x / x, a.y / x}; } int N, Log, home[MAXN]; point tmp[MAXN]; void FFTinit() { for (int i = 0; i < N; i++) { int tmp = i, ans = 0; for (int j = 1; j <= Log; j++) { ans <<= 1; ans += tmp & 1; tmp >>= 1; } home[i] = ans; } } void FFT(point *a, int mode) { for (int i = 0; i < N; i++) if (home[i] < i) swap(a[i], a[home[i]]); for (int len = 2; len <= N; len <<= 1) { point delta = (point) {cosl(2 * pi / len * mode), sinl(2 * pi / len * mode)}; for (int i = 0; i < N; i += len) { point now = (point) {1, 0}; for (int j = i, k = i + len / 2; k < i + len; j++, k++) { point tmp = a[j]; point tnp = a[k] * now; a[j] = tmp + tnp; a[k] = tmp - tnp; now = now * delta; } } } if (mode == -1) { for (int i = 0; i < N; i++) a[i] = a[i] / N; } } void times(int *a, int *b, int *c, int P, int limit) { N = 1, Log = 0; while (N <= 2 * limit) { N <<= 1; Log++; } static point ax[MAXN], ay[MAXN]; static point bx[MAXN], by[MAXN]; for (int i = 0; i <= limit; i++) { ax[i] = (point) {a[i] & 32767, 0}; ay[i] = (point) {a[i] >> 15, 0}; bx[i] = (point) {b[i] & 32767, 0}; by[i] = (point) {b[i] >> 15, 0}; } for (int i = limit + 1; i < N; i++) { ax[i] = (point) {0, 0}; ay[i] = (point) {0, 0}; bx[i] = (point) {0, 0}; by[i] = (point) {0, 0}; } FFTinit(); FFT(ax, 1), FFT(ay, 1), FFT(bx, 1), FFT(by, 1); static point x[MAXN], y[MAXN], z[MAXN]; for (int i = 0; i < N; i++) { x[i] = ax[i] * bx[i]; y[i] = ax[i] * by[i] + ay[i] * bx[i]; z[i] = ay[i] * by[i]; } FFT(x, -1), FFT(y, -1), FFT(z, -1); auto num = [&] (point x) { return (long long) (x.x + 0.5) % P; }; for (int i = 0; i < N; i++) { int res = num(z[i]); res = (32768ll * res + num(y[i])) % P; res = (32768ll * res + num(x[i])) % P; c[i] = res; } } } ll n; int k; int dp[MAXLOG][MAXN]; int fac[MAXN], inv[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; } void init(int n) { fac[0] = 1; for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % P; inv[n] = power(fac[n], P - 2); for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1ll) % P; } int main() { read(n), read(k); if (n > k) { puts("0"); return 0; } init(k); for (int i = 1; i <= k; i++) dp[0][i] = inv[i]; for (int p = 1; p < MAXLOG; p++) { static int a[MAXN], b[MAXN], c[MAXN]; for (int i = 0; i <= k; i++) { a[i] = 1ll * dp[p - 1][i] * power(2, i * (1 << (p - 1))) % P; b[i] = dp[p - 1][i]; } AnyModuloFFT :: times(a, b, c, P, k); for (int i = 0; i <= k; i++) dp[p][i] = c[i]; } static int curr[MAXN]; curr[0] = 1; for (int p = MAXLOG - 1; p >= 0; p--) { if ((n & (1 << p)) == 0) continue; static int a[MAXN], b[MAXN], c[MAXN]; for (int i = 0; i <= k; i++) { a[i] = 1ll * curr[i] * power(2, i * (1 << p)) % P; b[i] = dp[p][i]; } AnyModuloFFT :: times(a, b, c, P, k); for (int i = 0; i <= k; i++) curr[i] = c[i]; } int ans = 0; for (int i = n; i <= k; i++) ans = (ans + 1ll * curr[i] * inv[k - i]) % P; writeln(1ll * ans * fac[k] % P); return 0; }