Rokua P4233: Shot pill Notes (counting principle + tolerant principle + polynomial seeking inverse/sub-treatment NTT)

Topic Portal: https://www.luogu.org/problemnew/show/P4233

Topic Analysis: This problem is I play the game when I saw, but at that time did not want to come out, even some points are not bothered to write Qaq. Later I analyzed why I always can not think of the NTT's topic, found that it is their own count, tolerance, DP learning too bad, the province before the game or to find a time to fill a supplement.

Others have written a very detailed explanation, I will not be wordy: the official portal

I add: This question and BZOJ3456 city planning that the problem is very similar, but this problem is enumerated in the topology map of the smallest of the strong connected components, and finally introduced a similar formula:

G[N]=∑K=1NF[K]∗G[N−K] G [n] =∑k = 1 n F [k]∗g [n−k] g[n]=\sum_{k=1}^{n}f[k]*g[n-k]

Known g, requires F. Then there is a new posture of polynomial inversion: Notice that this is probably a f*g=g form, and that f is a meta-function. But the Nether of K here is 1. In order for the Nether of K to become 0, we force the f[0]=0 to be made, then a (x) =g (x) (x>0), a (0) =0 A (x) = G (x) (x > 0), a (0) = 0 A (x) =g (x) (x>0), a (0) = 0, So there is A=f*g (G[0] Originally 1, but because F[0] is forced to 0, so a[0] must be 0). Then use polynomial inversion, time complexity O (Nlog (n)) O (n log (n)) O (N\log (n)).

CODE:

#include <iostream> #include <string> #include <cstring> #include <cmath> #include <cstdio

> #include <cstdlib> #include <stdio.h> #include <algorithm> using namespace std;
const int maxn=1000000;
Const long Long m=998244353;
Const long Long g=3;

typedef long Long LL;
LL A[MAXN];

LL B[MAXN];
int REV[MAXN];

int N,LG;
LL F[MAXN];

LL G[MAXN];
LL FAC[MAXN];
LL NFAC[MAXN];

int n;
    ll Pow (ll X,ll y) {if (!y) return 1LL;
    LL Temp=pow (x,y>>1);
    temp=temp*temp%m;
    if (y&1) temp=temp*x%m;
return temp;

    } void DFT (LL *a,int f) {for (int i=0; i<n; i++) if (I<rev[i]) swap (a[i],a[rev[i]);
        for (int len=2; len<=n; len<<=1) {int mid= (len>>1);
        LL E=pow (g, (M-1)/len);

        if (f==-1) E=pow (e,m-2);
            for (ll *p=a; p!=a+n; p+=len) {ll wn=1;
           for (int i=0; i<mid; i++) {LL temp=wn*p[mid+i]%m;     p[mid+i]= (p[i]-temp+m)%M;
                p[i]= (p[i]+temp)%M;
            wn=wn*e%m;
        }}}} void NTT (bool rev) {for (int i=0; i<n; i++) {rev[i]=0;
    for (int j=0; j<lg; J + +) if (i& (1<<j)) rev[i]|= (1<< (lg-j-1));
    } DFT (a,1);
    DFT (b,1);
    if (rev.) for (int i=0; i<n; i++) a[i]=a[i]*a[i]%m;
    for (int i=0; i<n; i++) a[i]=a[i]*b[i]%m;

    DFT (a,-1);
    LL Inv=pow (n,m-2);
for (int i=0; i<n; i++) a[i]=a[i]*inv%m;
    } void Poly_rev (int m) {if (m==1) F[0]=pow (g[0],m-2);

        else {Poly_rev (m>>1);
        n=2*m,lg=0;
        while ((1&LT;&LT;LG)!=n) lg++;
        for (int i=0; i<m; i++) a[i]=f[i],b[i]=g[i];
        for (int i=m; i<n; i++) a[i]=b[i]=0;

        NTT (TRUE);
    for (int i=0; i<m; i++) f[i]=2ll*f[i]%m,f[i]= (f[i]-a[i]+m)%M;
    }} int main () {freopen ("4233.in", "R", stdin);

    Freopen ("4233.out", "w", stdout); scanf ("%D ", &n);
    N=1;
    while (n<=n+2) n<<=1;
    Fac[0]=1;
    for (LL i=1; i<n; i++) fac[i]=fac[i-1]*i%m;
    for (int i=0; i<n; i++) Nfac[i]=pow (fac[i],m-2);
    for (LL i=0; i<n; i++) G[i]=pow (2LL, i* (I-1LL)/2ll) *nfac[i]%m;

    Poly_rev (N);
    n=1,lg=0;
    while (n<2*n+2) n<<=1,lg++;
    for (int i=0; i<=n; i++) a[i]=f[i],b[i]=g[i];
    for (int i=n+1; i<n; i++) a[i]=b[i]=0;
    b[0]=0;
    NTT (FALSE);
        for (LL i=1; i<=n; i++) {a[i]=a[i]*fac[i]%m; if (!
        A[i]) printf (" -1\n");
            else {LL Ans=fac[i-1]*pow (2LL, Max (i* (I-1LL)/2ll-i,0ll))%M;
            Ans=ans*pow (a[i],m-2)%M;
        printf ("%i64d\n", ans);
}} return 0; }