「LOJ 556 Antileaf's Round」咱们去烧菜吧

「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:
\[ \prod_{1\leq k \leq m}\left(\sum_{0\leq i\leq b_k} x^{ia_k}\right) \\ =\prod_{1\leq k\leq m}\left(\dfrac{1-x^{a_k{(b_k+1)}}}{1-x^{a_k}}\right) \\ =\exp\left(\sum_{1\leq k\leq m}\ln(1-x^{a_k(b_k+1)})-\ln(1-x^{a_k})\right) \\ =\exp\left(\sum_{1\leq k\leq m}\sum_{j\geq1}\dfrac{x^{a_kj}-x^{a_k(b_k+1)j}}{j}\right) \\ \]


\[ c_k =\sum_{1\leq i\leq m} [a_i=k] \\d_k=\sum_{1\leq i\leq m}[a_i(b_i+1)=k] \\ A(x)=\sum_{k}(c_k-d_k)x^k \]
上面式子等价于
\[ =\exp\left(\sum_{j\geq1}\dfrac{1}{j}\sum_{k}(c_k-d_k)x^{jk}\right) \\ =\exp\left(\sum_{j\geq1}\dfrac{1}{j}A(x^j)\right) \]
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 \(\lfloor\dfrac{n}{j}\rfloor\) 个,可以 \(\mathcal O(n\log n)\) 预处理出
\[ \sum_{j\geq1}\dfrac{1}{j}A(x^j) \]
然后跑一遍多项式 \(\exp\) 就好了,复杂度还是 \(\mathcal O(n\log n)\)

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\n", g[i]);
    return 0;
}
posted @ 2019-04-12 08:16 Joyemang33 阅读(...) 评论(...) 编辑 收藏