「LOJ 556 Antileaf's Round」咱们去烧菜吧
最近在看 jcvb 的生成函数课件，顺便切一切上面讲到的内容的板子题，这个题和课件上举例的背包计数基本一样。
解题思路
首先列出答案的生成函数：
[prod_{1leq k leq m}left(sum_{0leq ileq b_k} x^{ia_k}
ight) \
=prod_{1leq kleq m}left(dfrac{1-x^{a_k{(b_k+1)}}}{1-x^{a_k}}
ight) \
=expleft(sum_{1leq kleq m}ln(1-x^{a_k(b_k+1)})-ln(1-x^{a_k})
ight) \
=expleft(sum_{1leq kleq m}sum_{jgeq1}dfrac{x^{a_kj}-x^{a_k(b_k+1)j}}{j}
ight) \
]
令
[c_k =sum_{1leq ileq m} [a_i=k]
\d_k=sum_{1leq ileq m}[a_i(b_i+1)=k] \
A(x)=sum_{k}(c_k-d_k)x^k
]
上面式子等价于
[=expleft(sum_{jgeq1}dfrac{1}{j}sum_{k}(c_k-d_k)x^{jk}
ight) \
=expleft(sum_{jgeq1}dfrac{1}{j}A(x^j)
ight)
]
因为过程始终在模 (x^{n+1}) 次下进行，所以 (A(x^j)) 有用的项只有 (lfloordfrac{n}{j} floor) 个，可以 (mathcal O(nlog n)) 预处理出
[sum_{jgeq1}dfrac{1}{j}A(x^j)
]
然后跑一遍多项式 (exp) 就好了，复杂度还是 (mathcal O(nlog n)) 。
这个题数据有问题，当 (a_i = 0) 的时候，第一步不能等比数列求和，然而通过此题需要无视 (a_i = 0) 的物品，甚至还有 (a_i=0,b_i=0) 的情况，所以下面贴一个AC代码，但是实际山来说不是正确的
code
/*program by mangoyang*/
#include<bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = (1 << 22) + 5, P = 998244353, G = 3;
namespace poly{
int rev[N], W[N], invW[N], len, lg;
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % P)
if(b & 1) ans = 1ll * ans * a % P;
return ans;
}
inline void init(){
for(int k = 2; k < N; k <<= 1)
W[k] = Pow(G, (P - 1) / k), invW[k] = Pow(W[k], P - 2);
}
inline void timesinit(int lenth){
for(len = 1, lg = 0; len <= lenth; len <<= 1, lg++);
for(int i = 0; i < len; i++)
rev[i] = (rev[i>>1] >> 1) | ((i & 1) << (lg - 1));
}
inline void DFT(int *a, int sgn){
for(int i = 0; i < len; i++) if(i < rev[i]) swap(a[i], a[rev[i]]);
for(int k = 2; k <= len; k <<= 1){
int w = ~sgn ? W[k] : invW[k];
for(int i = 0; i < len; i += k){
int now = 1;
for(int j = i; j < i + (k >> 1); j++){
int x = a[j], y = 1ll * a[j+(k>>1)] * now % P;
a[j] = (x + y) % P, a[j+(k>>1)] = (x - y + P) % P;
now = 1ll * now * w % P;
}
}
}
if(sgn == -1){
int Inv = Pow(len, P - 2);
for(int i = 0; i < len; i++) a[i] = 1ll * a[i] * Inv % P;
}
}
inline void getinv(int *a, int *b, int n){
static int tmp[N];
if(n == 1) return (void) (b[0] = Pow(a[0], P - 2));
getinv(a, b, (n + 1) / 2);
timesinit(n * 2 - 1);
for(int i = 0; i < len; i++) tmp[i] = i < n ? a[i] : 0;
DFT(tmp, 1), DFT(b, 1);
for(int i = 0; i < len; i++)
b[i] = 1ll * (2 - 1ll * tmp[i] * b[i] % P + P) % P * b[i] % P;
DFT(b, -1);
for(int i = n; i < len; i++) b[i] = 0;
for(int i = 0; i < len; i++) tmp[i] = 0;
}
inline void getsqrt(int *a, int *b, int n){
static int tmp1[N], tmp2[N];
if(n == 1) return (void) (b[0] = 1);
getsqrt(a, b, (n + 1) / 2);
for(int i = 0; i < n; i++) tmp1[i] = a[i];
getinv(b, tmp2, n), timesinit(n * 2 - 1);
DFT(tmp1, 1), DFT(tmp2, 1);
for(int i = 0; i < len; i++) tmp1[i] = 1ll * tmp1[i] * tmp2[i] % P;
DFT(tmp1, -1);
for(int i = 0; i < len; i++)
b[i] = 1ll * (b[i] + tmp1[i]) % P * Pow(2, P - 2) % P;
for(int i = n; i < len; i++) b[i] = 0;
for(int i = 0; i < len; i++) tmp1[i] = tmp2[i] = 0;
}
inline void getln(int *a, int *b, int n){
static int tmp[N];
getinv(a, b, n), timesinit(n * 2 - 1);
for(int i = 1; i < n; i++) tmp[i-1] = 1ll * a[i] * i % P;
DFT(tmp, 1), DFT(b, 1);
for(int i = 0; i < len; i++) b[i] = 1ll * tmp[i] * b[i] % P;
DFT(b, -1);
for(int i = len - 1; i; i--) b[i] = 1ll * b[i-1] * Pow(i, P - 2) % P;
b[0] = 0;
for(int i = n; i < len; i++) b[i] = 0;
for(int i = 0; i < len; i++) tmp[i] = 0;
}
inline void getexp(int *a, int *b, int n){
static int tmp[N];
if(n == 1) return (void) (b[0] = 1);
getexp(a, b, (n + 1) / 2);
getln(b, tmp, n), timesinit(n * 2 - 1);
for(int i = 0; i < n; i++) tmp[i] = (!i - tmp[i] + a[i] + P) % P;
DFT(tmp, 1), DFT(b, 1);
for(int i = 0; i < len; i++) b[i] = 1ll * b[i] * tmp[i] % P;
DFT(b, -1);
for(int i = n; i < len; i++) b[i] = 0;
for(int i = 0; i < len; i++) tmp[i] = 0;
}
inline void power(int *a, int *b, int n, int m, ll k){
static int tmp[N];
for(int i = 0; i < m; i++) b[i] = 0;
int fir = -1;
for(int i = 0; i < n; i++) if(a[i]){ fir = i; break; }
if(fir && k >= m) return;
if(fir == -1 || 1ll * fir * k >= m) return;
for(int i = fir; i < n; i++) b[i-fir] = a[i];
for(int i = 0; i < n - fir; i++)
b[i] = 1ll * b[i] * Pow(a[fir], P - 2) % P;
getln(b, tmp, m);
for(int i = 0; i < m; i++)
b[i] = 1ll * tmp[i] * (k % P) % P, tmp[i] = 0;
getexp(b, tmp, m);
for(int i = m; i >= fir * k; i--)
b[i] = 1ll * tmp[i-fir*k] * Pow(a[fir], k % (P - 1)) % P;
for(int i = 0; i < fir * k; i++) b[i] = 0;
for(int i = 0; i < m; i++) tmp[i] = 0;
}
}
using poly::Pow;
using poly::DFT;
using poly::timesinit;
int a[N], b[N], A[N], f[N], g[N], n, m;
int main(){
poly::init(), read(n), read(m);
for(int i = 1; i <= m; i++){
read(a[i]), read(b[i]);
if(!b[i]) b[i] = n / a[i];
if(a[i] <= n) A[a[i]]++;
if(1ll * a[i] * (b[i] + 1) <= n) A[a[i]*(b[i]+1)]--;
}
for(int i = 0; i <= n; i++) (A[i] += P) %= P;
for(int j = 1; j <= n; j++){
int Inv = Pow(j, P - 2);
for(int i = 1; i <= n / j; i++)
(f[i*j] += 1ll * A[i] * Inv % P) %= P;
}
poly::getexp(f, g, n + 1);
for(int i = 1; i <= n; i++) printf("%d
", g[i]);
return 0;
}