[FWT] 51Nod 算法马拉松26 A A国的贸易

来源：互联网发布：centos搭建lnmp 编辑：程序博客网时间：2024/06/06 15:01

相当于每次每个点会变成自己和与自己相差一个二进制位的数的和
直接FWT 快速幂

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline void write(int x){  if (x>=10) write(x/10);  putchar(x%10+'0');}const int N=1<<20;const int P=1e9+7;const int INV2=(P+1)>>1;inline ll Pow(ll a,int b){  ll ret=1;  for (;b;b>>=1,a=a*a%P)    if (b&1)      ret=ret*a%P;  return ret;}int n,t;int a[N],b[N];inline void FWT(int *a,int n,int r){  for (int i=1;i<n;i<<=1)    for (int j=0;j<n;j+=(i<<1))      for (int k=0;k<i;k++){    int x=a[j+k],y=a[j+k+i];    if (r) a[j+k]=(x+y)%P,a[j+k+i]=(x+P-y)%P;    else a[j+k]=(ll)(x+y)*INV2%P,a[j+k+i]=(ll)(x+P-y)*INV2%P;      }}int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); read(t);  for (int i=0;i<(1<<n);i++) read(a[i]);  for (int i=0;i<n;i++) b[1<<i]=1; b[0]=1;  FWT(a,1<<n,1);  FWT(b,1<<n,1);  for (int i=0;i<(1<<n);i++) a[i]=a[i]*Pow(b[i],t)%P;  FWT(a,1<<n,0);  for (int i=0;i<(1<<n);i++)    write(a[i]),putchar(' ');  return 0;}

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