cf453D(FWT+快速乘)

题目链接

https://codeforces.com/contest/453/problem/D

题意

给定 $b$ 、$t$ 和长度为 $2^m$ 的数组 $e0$ ,已知 $e_i[u]=\sum{v}e_{i-1}b[|u\oplus v|]$ ,求 $e_t$

题解

直接把 $b$ 扩展成 $2^m$ 的数组,然后就是 $e_0$ 和 $b^n$ 做卷积。。

然后模数不是质数,做 $IFWT$ 的时候不能直接除以 $2^m$ ,所以要将模数乘上 $2^m$ ,然后发现会爆 longlong ,需要快速乘。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<cstdlib>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define mid (x+y>>1)
#define sqr(x) ((x)*(x))
#define NM 1100005
#define nm 400005
using namespace std;
const double pi=acos(-1);
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}


int n,m;
ll _t,inf;
ll a[NM],b[NM],c[NM];
inline ll plu(long long x,long long y){
ll tmp=(x*y-(ll)((long double)x/inf*y+1.0e-8)*inf);
return tmp<0 ? tmp+inf : tmp;
}

ll qpow(ll x,ll t){
ll s=1;
for(;t;t>>=1,x=plu(x,x))if(t&1)s=plu(s,x);
return s;
}


void fwt(ll*a){
for(int k=1;k<n;k<<=1)
for(int i=0;i<n;i+=k<<1)
for(int j=0;j<k;j++){
ll x=a[i+j],y=a[i+j+k];
a[i+j]=(x+y)%inf;
a[i+j+k]=(x-y+inf)%inf;
}
}

int main(){
m=read();n=succ(m);_t=read();inf=read()*n;
inc(i,0,n-1)a[i]=read()%inf;
inc(i,0,m)b[i]=read()%inf;
inc(i,0,n-1)c[i]=b[__builtin_popcount(i)];
fwt(c);fwt(a);
inc(i,0,n-1)a[i]=plu(a[i],qpow(c[i],_t));
fwt(a);
inc(i,0,n-1)printf("%lld\n",a[i]/n);
return 0;
}