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$b_i=\sum\limits_{j=i-kx}a_j$, $(0\le x,1\le j\le i\le n)$
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$(\sum x^{ki})^n=\sum\binom{n-1+i}{i}x^{ki}$, 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\n", ans); } }