题面

设\(a\)的递推公式为

\[a_i=\sum_ja_jb[count(i\oplus j)]\]

其中\(\oplus\)为异或，\(count(i)\)表示\(i\)的二进制中\(1\)的个数

给出\(a_0,b\)，求\(a_t\)，\(t\leq 10^{18}\)

题解

如果我们定义\(c_i=b[count(i)]\)

这显然就是个异或卷积了……因为要卷\(t\)次，所以点值表示乘起来的时候要把\(c_i\)快速幂一下

然而有个尴尬的问题就是这里的模数可能是偶数……那么我们\(IDFT\)的时候\(2\)显然没有逆元啊……

解决方法是把模数乘上\(lim\)（即\(fwt\)的数组长度），那么最后\(IDFT\)之后把所有数对\(lim\)下去整就行了

记得得用快速乘

//minamoto#include
    
     #define R register#define ll long long#define dd long double#define fp(i,a,b) for(R int i=(a),I=(b)+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    R int res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}ll readll(){    R ll res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}const int N=(1<<20)+5;ll P;inline ll add(R ll x,R ll y){return x+y>=P?x+y-P:x+y;}inline ll dec(R ll x,R ll y){return x-y<0?x-y+P:x-y;}inline ll mul(R ll x,R ll y){return x*y-(ll)((dd)x/P*y)*P;}inline ll ksm(R ll x,R ll y){    ll res=1;    for(;y;y>>=1,x=mul(x,x))y&1?res=mul(res,x):0;    return res;}void Fwt(ll *A,int lim,int ty){    ll t;    for(R int mid=1;mid
      
       <<=1)        for(R int j=0;j
       
        <<1))            fp(k,0,mid-1)                A[j+k+mid]=dec(A[j+k],t=A[j+k+mid]),                A[j+k]=add(A[j+k],t);    if(!ty)fp(i,0,lim-1)A[i]/=lim;}int n,lim;ll t,a[N],c[N],sz[N],b[25];int main(){    n=read(),t=readll(),P=read(),lim=(1<
        
         >1]+(i&1)];    Fwt(a,lim,1),Fwt(c,lim,1);    fp(i,0,lim-1)a[i]=mul(a[i],ksm(c[i],t));    Fwt(a,lim,0);    fp(i,0,lim-1)printf("%I64d\n",a[i]);    return 0;}