[luogu4389]付公主的背包(FFT)

完全背包方案计数问题的FFT优化。
首先写成生成函数的形式：对重量为V的背包，它的生成函数为$\sum\limits_{i=0}^{+\infty}x^{Vi}=\frac{1}{1-x^{V}}$
于是答案就是$\prod \frac{1}{1-x^{V_k}}$。
直接做显然会超时，考虑使用ln将乘法变为加法。
https://www.cnblogs.com/cjyyb/p/10132855.html

 1 #include<cmath>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #define mem(a) memset(a,0,sizeof(a))
 6 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 7 using namespace std;
 8
 9 const int N=500010,mod=998244353,inv2=(mod+1)/2;
10 int n,m,v,cnt[N],rev[N],inv[N],X[N],Y[N],A[N],D[N],E[N],F[N];
11
12 void Print(int a[],int n=::n){ for (int i=0; i<n; i++) printf("%d ",a[i]); puts(""); }
13
14 int ksm(int a,int b){
15     int res=1;
16     for (; b; a=1ll*a*a%mod,b>>=1)
17         if (b & 1) res=1ll*res*a%mod;
18     return res;
19 }
20
21 void NTT(int a[],int n,bool f){
22     for (int i=0; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
23     for (int i=1; i<n; i<<=1){
24         int wn=ksm(3,f ? (mod-1)/(i<<1) : (mod-1)-(mod-1)/(i<<1));
25         for (int p=i<<1,j=0; j<n; j+=p){
26             int w=1;
27             for (int k=0; k<i; k++,w=1ll*w*wn%mod){
28                 int x=a[j+k],y=1ll*w*a[i+j+k]%mod;
29                 a[j+k]=(x+y)%mod; a[i+j+k]=(x-y+mod)%mod;
30             }
31         }
32     }
33     if (f) return;
34     int inv=ksm(n,mod-2);
35     for (int i=0; i<n; i++) a[i]=1ll*a[i]*inv%mod;
36 }
37
38 void mul(int a[],int b[],int l){
39     int n=1,L=0;
40     for (; n<(l<<1); n<<=1) L++;
41     for (int i=0; i<n; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
42     NTT(a,n,1); NTT(b,n,1);
43     for (int i=0; i<n; i++) a[i]=1ll*a[i]*b[i]%mod;
44     NTT(a,n,0); NTT(b,n,0);
45 }
46
47 void Inv(int a[],int b[],int l){
48     if (l==1){ b[0]=ksm(a[0],mod-2); return; }
49     Inv(a,b,l>>1); int n=1,L=0;
50     for (; n<(l<<1); n<<=1) L++;
51     for (int i=0; i<n; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
52     for (int i=0; i<l; i++) A[i]=a[i];
53     NTT(A,n,1); NTT(b,n,1);
54     for (int i=0; i<n; i++) b[i]=1ll*b[i]*(2-1ll*A[i]*b[i]%mod+mod)%mod;
55     NTT(b,n,0);
56     for (int i=l; i<n; i++) b[i]=0;
57     for (int i=0; i<n; i++) A[i]=0;
58 }
59
60 void Deri(int a[],int b[],int l){
61     for (int i=1; i<l; i++) b[i-1]=1ll*i*a[i]%mod;
62 }
63
64 void Inte(int a[],int b[],int l){
65     for (int i=1; i<l; i++) b[i]=1ll*a[i-1]*inv[i]%mod; b[0]=0;
66 }
67
68 void Ln(int a[],int b[],int l){
69     Deri(a,D,l); Inv(a,E,l); mul(D,E,l); Inte(D,b,l);
70     for (int i=0; i<(l<<1); i++) D[i]=E[i]=0;
71 }
72
73 void Exp(int a[],int b[],int l){
74     if (l==1){ b[0]=1; return; }
75     Exp(a,b,l>>1); Ln(b,F,l); int n=1,L=0;
76     for (; n<(l<<1); n<<=1) L++;
77     for (int i=0; i<n; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
78     for (int i=0; i<l; i++) F[i]=(-F[i]+a[i]+mod)%mod; F[0]=(F[0]+1)%mod;
79     NTT(F,n,1); NTT(b,n,1);
80     for (int i=0; i<n; i++) b[i]=1ll*b[i]*F[i]%mod;
81     NTT(b,n,0);
82     for (int i=l; i<n; i++) b[i]=0;
83     for (int i=0; i<n; i++) F[i]=0;
84 }
85
86 int main(){
87     freopen("4389.in","r",stdin);
88     freopen("4389.out","w",stdout);
89     scanf("%d%d",&n,&m);
90     int l=1; for (; l<=m; l<<=1); inv[0]=inv[1]=1;
91     rep(i,2,l) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
92     rep(i,1,n) scanf("%d",&v),cnt[v]++;
93     rep(i,1,m) if (cnt[i]) for (int j=1; j*i<=m; j++) X[j*i]=(X[j*i]+1ll*cnt[i]*inv[j])%mod;
94     Exp(X,Y,l);
95     rep(i,1,m) printf("%d\n",Y[i]);
96     return 0;
97 }

原文地址：https://www.cnblogs.com/HocRiser/p/10374025.html