F(n)实际上是一个等比数列的和，将它从递推式转变为通项公式(a^n-1)/(a-1)，这里只需要确定n就可以。
题目要求选取k大小的所有子集的答案求和，可以先求a^n部分的和，把它写成母函数的形式(x+a^s1)(x+a^s2)...(x+a^sn)，这样不断的分成两半，分别求出，然后再用fft卷积合并这两半，这一部分的和就是x^(n-k)的系数。
分母的a-1实际上还是a-1，分子的-1部分要变成C(n,k)
由于之前没怎么写过，fft的板子代码借鉴群里大佬的。
#include #include #include <string> #include #include #include #include #include #include <set> #include #include #include #include const int maxn = 1e5 + 10; const int mod = 1e5 + 3; using namespace std; typedef long long ll; const double PI = acos(-1.0); namespace fft { struct num { double x, y; num() { x = y = 0; } num(double x, double y) : x(x), y(y) {} }; inline num operator+(num a, num b) { return num(a.x + b.x, a.y + b.y); } inline num operator-(num a, num b) { return num(a.x - b.x, a.y - b.y); } inline num operator*(num a, num b) { return num(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); } inline num conj(num a) { return num(a.x, -a.y); } int base = 1; vector roots = {{0, 0}, {1, 0}}; vector<int> rev = {0, 1}; const double PI = acosl(-1.0); void ensure_base(int nbase) { if (nbase <= base) return; rev.resize(1 << nbase); for (int i = 0; i < (1 << nbase); i++) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (nbase - 1)); roots.resize(1 << nbase); while (base < nbase) { double angle = 2 * PI / (1 << (base + 1)); for (int i = 1 << (base - 1); i < (1 << base); i++) { roots[i << 1] = roots[i]; double angle_i = angle * (2 * i + 1 - (1 << base)); roots[(i << 1) + 1] = num(cos(angle_i), sin(angle_i)); } base++; } } void fft(vector &a, int n = -1) { if (n == -1) n = a.size(); assert((n & (n - 1)) == 0); int zeros = __builtin_ctz(n); ensure_base(zeros); int shift = base - zeros; for (int i = 0; i < n; i++) if (i < (rev[i] >> shift)) swap(a[i], a[rev[i] >> shift]); for (int k = 1; k < n; k <<= 1) { for (int i = 0; i < n; i += 2 * k) { for (int j = 0; j < k; j++) { num z = a[i + j + k] * roots[j + k]; a[i + j + k] = a[i + j] - z; a[i + j] = a[i + j] + z; } } } } vector fa, fb; vector<int> multiply_mod(vector<int> &a, vector<int> &b, int m, int eq = 0) { int need = a.size() + b.size() - 1; int nbase = 0; while ((1 << nbase) < need) nbase++; ensure_base(nbase); int sz = 1 << nbase; if (sz > (int)fa.size()) fa.resize(sz); for (int i = 0; i < (int)a.size(); i++) { int x = (a[i] % m + m) % m; fa[i] = num(x & ((1 << 15) - 1), x >> 15); } fill(fa.begin() + a.size(), fa.begin() + sz, num{0, 0}); fft(fa, sz); if (sz > (int)fb.size()) fb.resize(sz); if (eq) copy(fa.begin(), fa.begin() + sz, fb.begin()); else { for (int i = 0; i < (int)b.size(); i++) { int x = (b[i] % m + m) % m; fb[i] = num(x & ((1 << 15) - 1), x >> 15); } fill(fb.begin() + b.size(), fb.begin() + sz, num{0, 0}); fft(fb, sz); } double ratio = 0.25 / sz; num r2(0, -1), r3(ratio, 0), r4(0, -ratio), r5(0, 1); for (int i = 0; i <= (sz >> 1); i++) { int j = (sz - i) & (sz - 1); num a1 = (fa[i] + conj(fa[j])); num a2 = (fa[i] - conj(fa[j])) * r2; num b1 = (fb[i] + conj(fb[j])) * r3; num b2 = (fb[i] - conj(fb[j])) * r4; if (i != j) { num c1 = (fa[j] + conj(fa[i])); num c2 = (fa[j] - conj(fa[i])) * r2; num d1 = (fb[j] + conj(fb[i])) * r3; num d2 = (fb[j] - conj(fb[i])) * r4; fa[i] = c1 * d1 + c2 * d2 * r5; fb[i] = c1 * d2 + c2 * d1; } fa[j] = a1 * b1 + a2 * b2 * r5; fb[j] = a1 * b2 + a2 * b1; } fft(fa, sz); fft(fb, sz); vector<int> res(need); for (int i = 0; i < need; i++) { ll aa = fa[i].x + 0.5; ll bb = fb[i].x + 0.5; ll cc = fa[i].y + 0.5; res[i] = (aa + ((bb % m) << 15) + ((cc % m) << 30)) % m; } return res; } vector<int> square_mod(vector<int> &a, int m) { return multiply_mod(a, a, m, 1); } }; // namespace fft inline int quick(int a, int b, int m) { int ans = 1; while (b) { if (b & 1) { ans = (1LL * a * ans) % m; } a = (1LL * a * a) % m; b >>= 1; } return ans; } vector<int> v[maxn]; inline ll inverse(ll a, ll p) { return quick(a, p - 2, p); } vector<int> solve(int l, int r) { if (l == r) return v[l]; int mid = (l + r) >> 1; vector<int> v1 = (solve(l, mid)); vector<int> v2 = solve(mid + 1, r); return fft::multiply_mod(v1, v2, mod); } int comb[maxn]; int main() { int n, a, q; scanf("%d%d%d", &n, &a, &q); comb[0] = 1; for (int i = 1; i <= n; i++) { comb[i] = 1LL * comb[i - 1] * (n + 1 - i) * inverse(i, mod) % mod; } for (int i = 1, s; i <= n; i++) { scanf("%d", &s); v[i].push_back(quick(a % mod, s, mod)); v[i].push_back(1); } vector<int> ans = solve(1, n); while (q--) { int k; scanf("%d", &k); int cnt = 1LL * (ans[n - k] - comb[k]) * inverse(a - 1, mod) % mod; if (cnt < 0) cnt += mod; printf("%d\n", cnt); } return 0; }
转载于:https://www.cnblogs.com/hyfer/p/11492428.html
