1. 程式人生 > >[多項式ln][多項式exp][揹包DP][生成函式] LOJ #556. 咱們去燒菜吧

[多項式ln][多項式exp][揹包DP][生成函式] LOJ #556. 咱們去燒菜吧

Solution

就是個揹包DP咯。
答案長這個樣子

ans=[xn]i=1m1xaibi+ai1xai考慮取個lnlnans=i=1m(ln(1xaibi+ai)ln(1xai))根據ln(1xk)=i1xkii展開,用一個桶存一下每個式子的貢獻,展開的複雜度就是O(nlnn)的。
最後求個exp就好了。O(nlogn)
#include <bits/stdc++.h>
#define show(x) cerr << #x << " = " << x << endl using namespace std; typedef long long ll; typedef pair<int, int> pairs; const int N = 303030; const int MOD = 998244353; inline char get(void) { static char buf[100000], *S = buf, *T = buf; if (S == T) { T = (S = buf) + fread(buf, 1
, 100000, stdin); if (S == T) return EOF; } return *S++; } template<typename T> inline void read(T &x) { static char c; x = 0; int sgn = 0; for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1; for (; c >= '0' && c <= '9'; c = get()) x = x * 10
+ c - '0'; if (sgn) x = -x; } inline int pwr(int a, int b) { int c = 1; while (b) { if (b & 1) c = (ll)c * a % MOD; b >>= 1; a = (ll)a * a % MOD; } return c; } inline int ivs(int x) { return pwr(x, MOD - 2); } inline int sum(int a, int b) { a += b; return a >= MOD ? a - MOD : a; } inline int sub(int a, int b) { return a < b ? a - b + MOD : a - b; } inline void add(int &x, int a) { x = sum(x, a); } namespace FNT { const int MAXN = 303030; int ww[MAXN], iw[MAXN]; int rev[MAXN]; int num; inline void pre(int n) { num = n; int g = pwr(3, (MOD - 1) / n); ww[0] = iw[0] = 1; for (int i = 1; i < num; i++) iw[n - i] = ww[i] = (ll)ww[i - 1] * g % MOD; } inline void fnt(int *a, int n, int f) { static int x, y, *w; w = (f == 1) ? ww : iw; for (int i = 0; i < n; i++) if (rev[i] > i) swap(a[rev[i]], a[i]); for (int i = 1; i < n; i <<= 1) for (int j = 0; j < n; j += (i << 1)) for (int k = 0; k < i; k++) { x = a[j + k]; y = (ll)a[j + k + i] * w[num / (i << 1) * k] % MOD; a[j + k] = sum(x, y); a[j + k + i] = sub(x, y); } if (f == -1){ int in = ivs(n); for (int i = 0; i < n; i++) a[i] = (ll)a[i] * in % MOD; } } } int inv[N]; inline void pre(int n) { inv[1] = 1; for (int i = 2; i <= n; i++) inv[i] = (ll)(MOD - MOD / i) * inv[MOD % i] % MOD; } void getInv(int *a, int *b, int n) { using namespace FNT; static int tmp[N]; if (n == 1) return (void)(b[0] = ivs(a[0])); getInv(a, b, n >> 1); for (int i = 0; i < n; i++) { tmp[i] = a[i]; tmp[i + n] = 0; } int L = 0; while (!(n >> L & 1)) L++; for (int i = 0; i < (n << 1); i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L); fnt(tmp, n << 1, 1); fnt(b, n << 1, 1); for (int i = 0; i < (n << 1); i++) tmp[i] = (ll)b[i] * sub(2, (ll)tmp[i] * b[i] % MOD) % MOD; fnt(tmp, n << 1, -1); for (int i = 0; i < n; i++) { b[i] = tmp[i]; b[n + i] = 0; } } inline void getLn(int *a, int *b, int n) { using namespace FNT; static int da[N], ia[N], tmp[N]; for (int i = 0; i < (n << 1); i++) tmp[i] = da[i] = ia[i] = 0; getInv(a, ia, n); for (int i = 1; i < n; i++) da[i - 1] = (ll)a[i] * i % MOD; int L = 0; while (!(n >> L & 1)) L++; for (int i = 0; i < (n << 1); i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L); fnt(da, n << 1, 1); fnt(ia, n << 1, 1); for (int i = 0; i < (n << 1); i++) tmp[i] = (ll)da[i] * ia[i] % MOD; fnt(tmp, n << 1, -1); b[0] = b[n] = 0; for (int i = 1; i < n; i++) { b[i] = (ll)tmp[i - 1] * inv[i] % MOD; b[n + i] = 0; } } inline void getExp(int *a, int *b, int n) { using namespace FNT; static int lb[N]; if (n == 1) return (void)(b[0] = 1); getExp(a, b, n >> 1); for (int i = 0; i < (n << 1); i++) lb[i] = 0; getLn(b, lb, n); for (int i = 0; i < n; i++) lb[i] = sub(a[i], lb[i]); lb[0] = sum(lb[0], 1); int L = 0; while (!(n >> L & 1)) L++; for (int i = 0; i < (n << 1); i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L); fnt(b, n << 1, 1); fnt(lb, n << 1, 1); for (int i = 0; i < (n << 1); i++) b[i] = (ll)b[i] * lb[i] % MOD; fnt(b, n << 1, -1); for (int i = 0; i < n; i++) b[i + n] = 0; } int n, m, l, x, y; int buc[N], f[N], g[N]; int main(void) { freopen("1.in", "r", stdin); freopen("1.out", "w", stdout); FNT::pre(1 << 18); pre(1 << 18); read(n); read(m); for (l = 1; l <= n; l <<= 1); for (int i = 1; i <= m; i++) { read(x); read(y); if (!x) continue; if (!y) y = n / x; if ((ll)x * y + x < l) buc[x * y + x]--; if (x < l) buc[x]++; } for (int i = 0; i < l; i++) buc[i] = sum(buc[i], MOD); for (int i = 1; i < l; i++) for (int j = i; j < l; j += i) f[j] = sum(f[j], (ll)buc[i] * inv[j / i] % MOD); getExp(f, g, l); for (int i = 1; i <= n; i++) printf("%d\n", g[i]); return 0; }