2019杭电多校一 L. Sequence (NTT)

大意: 给定序列$a$, 给定$m$个操作, 求最后序列每一项的值.

一共$3$种操作, 其中第$k$种操作将序列变为$b_i=sumlimits_{j=i-kx}a_j$, $(0le x,1le jle ile n)$

 

可以发现$sum b_ix^i=(sum a_i x^i)(sum x^{ki})$, 转化为求$(sum x^{ki})^{cnt}$

直接快速幂会$T$, 注意到$(sum x^{ki})^n=suminom{n-1+i}{i}x^{ki}$, 所以可以只求一次卷积

#include <iostream>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
using namespace std;
typedef long long ll;
typedef int* poly;
const int N = 2e6+10, P = 998244353, G = 3, Gi = 332748118;
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
int n,m,lim,l,A[N],B[N],R[N];
int fac[N],ifac[N];
void init(int n) {
    for (lim=1,l=0; lim<=n; lim<<=1,++l) ;
    REP(i,0,lim-1) R[i]=(R[i>>1]>>1)|((i&1)<<(l-1));
}
void NTT(poly J, int tp=1) {
    REP(i,0,lim-1) if (i<R[i]) swap(J[i],J[R[i]]);
    for (int j=1; j<lim; j<<=1) {
        ll T = qpow(tp==1?G:Gi,(P-1)/(j<<1));
        for (int k=0; k<lim; k+=j<<1) {
            ll t = 1;
            for (int l=0; l<j; ++l,t=t*T%P) {
                int y = t*J[k+j+l]%P;
                J[k+j+l] = (J[k+l]-y+P)%P;
                J[k+l] = (J[k+l]+y)%P;
            }
        }
    }
    if (tp==-1) {
        ll inv = qpow(lim, P-2);
        REP(i,0,lim-1) J[i]=(ll)inv*J[i]%P;
    }
}
poly mul(poly a, poly b) {
	init(n*2);
	REP(i,0,lim-1) A[i]=B[i]=0;
	REP(i,0,n-1) A[i]=a[i];
	REP(i,0,n-1) B[i]=b[i];
	NTT(A),NTT(B);
	poly c(new int[lim]);
	REP(i,0,lim-1) c[i]=(ll)A[i]*B[i]%P;
	NTT(c,-1);
	REP(i,0,lim-1) if (c[i]<0) c[i]+=P;
	return c;
}

int C(int n, int m) {
	if (n<m) return 0;
	return (ll)fac[n]*ifac[m]%P*ifac[n-m]%P;
}

int main() {
	fac[0]=ifac[0]=1;
	REP(i,1,N-1) fac[i]=(ll)fac[i-1]*i%P;
	ifac[N-1]=qpow(fac[N-1],P-2);
	PER(i,1,N-2) ifac[i]=(ll)ifac[i+1]*(i+1)%P;
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		poly a(new int[n]);
		REP(i,0,n-1) scanf("%d", a+i);
		int cnt[4]{};
		REP(i,1,m) {
			int c;
			scanf("%d", &c);
			++cnt[c];
		}
		REP(i,1,3) if (cnt[i]) {
			poly f(new int[n]());
			for (int j=0;j*i<n;++j) {
				f[j*i] = C(cnt[i]-1+j,j);
			}
			a = mul(a,f);
		}
		ll ans = 0;
		REP(i,0,n-1) ans ^= (i+1ll)*a[i];
		printf("%lld
", ans);
	}
}