牛客网暑期ACM多校训练营（第九场） A题 FWT

链接：https://www.nowcoder.com/acm/contest/147/A
来源：牛客网

Niuniu has recently learned how to use Gaussian elimination to solve systems of linear equations.

Given n and a[i], where n is a power of 2, let's consider an n x n matrix A.

The index of A[i][j] and a[i] are numbered from 0.
The element A[i][j] satisfies A[i][j] = a[i xor j],
https://en.wikipedia.org/wiki/Bitwise_operation#XOR

Let p = 1000000007.

Consider the equation 
A x = b (mod p)
where A is an n x n matrix, and x and b are both n x 1 row vector.

Given n, a[i], b[i], you need to solve the x.
For example, when n = 4, the equations look like
A[0][0]*x[0] + A[0][1]*x[1] + A[0][2]*x[2] + A[0][3]*x[3] = b[0] (mod p)
A[1][0]*x[0] + A[1][1]*x[1] + A[1][2]*x[2] + A[1][3]*x[3] = b[1] (mod p)
A[2][0]*x[0] + A[2][1]*x[1] + A[2][2]*x[2] + A[2][3]*x[3] = b[2] (mod p)
A[3][0]*x[0] + A[3][1]*x[1] + A[3][2]*x[2] + A[3][3]*x[3] = b[3] (mod p)
and the matrix A can be decided by the array a.

It is guaranteed that there is a unique solution x for these equations.

输入描述:

The first line contains an integer, which is n.
The second line contains n integers, which are the array a.
The third line contains n integers, which are the array b.

1 <= n <= 262144
p = 1000000007
0 <= a[i] < p
0 <= b[i] < p

输出描述:

The output should contains n lines.
The i-th(index from 0) line should contain x[i].
x[i] is an integer, and should satisfy 0 <= x[i] < p.

输入例子:

4
1 10 100 1000
1234 2143 3412 4321

输出例子:

4
3
2
1

-->

示例1

输入

4
1 10 100 1000
1234 2143 3412 4321

输出4

3
2
1

解析 　　给出        已知A b 求 x 

上式可以转化成           由于 i^j^j=i   

所以原式等式            直接套fwt_xor 板子

代码

 #include <bits/stdc++.h>
 using namespace std;
 #define rep(i,a,n) for (int i=a;i<n;i++)
 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 #define pb push_back
 #define mp make_pair
 #define all(x) (x).begin(),(x).end()
 #define fi first
 #define se second
 #define SZ(x) ((int)(x).size())
 typedef vector<int> VI;
 typedef long long ll;
 typedef pair<int,int> PII;
 const ll mod=;
 const int maxn = 3e6+;
 ll powmod(ll a,ll b) {ll res=;a%=mod; assert(b>=); for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
 // head

 int a[maxn],b[maxn];
 void FWT_or(int *a,int N,int opt)
 {
     for(int i=;i<N;i<<=)
         for(int p=i<<,j=;j<N;j+=p)
             for(int k=;k<i;++k)
                 if(opt==)a[i+j+k]=(a[j+k]+a[i+j+k])%mod;
                 else a[i+j+k]=(a[i+j+k]+mod-a[j+k])%mod;
 }
 void FWT_and(int *a,int N,int opt)
 {
     for(int i=;i<N;i<<=)
         for(int p=i<<,j=;j<N;j+=p)
             for(int k=;k<i;++k)
                 if(opt==)a[j+k]=(a[j+k]+a[i+j+k])%mod;
                 else a[j+k]=(a[j+k]+mod-a[i+j+k])%mod;
 }
 void FWT_xor(int *a,int N,int opt) //opt=1 正变换 opt=-1 逆变换
 {
     ll inv2=powmod(,mod-);
     for(int i=;i<N;i<<=)
         for(int p=i<<,j=;j<N;j+=p)
             for(int k=;k<i;++k)
             {
                 int X=a[j+k],Y=a[i+j+k];
                 a[j+k]=(X+Y)%mod;a[i+j+k]=(X+mod-Y)%mod;
                 if(opt==-)a[j+k]=1ll*a[j+k]*inv2%mod,a[i+j+k]=1ll*a[i+j+k]*inv2%mod;
             }
 }
 int main()
 {
     int n;
     scanf("%d",&n);
     for(int i=;i<n;i++)
         scanf("%d",&a[i]);
     for(int i=;i<n;i++)
         scanf("%d",&b[i]);
     FWT_xor(a,n,);FWT_xor(b,n,);
     for(int i=;i<n;i++)
     {
         b[i]=b[i]*powmod(a[i],mod-)%mod;   //  b/a
     }
     FWT_xor(b,n,-);   //再逆变换
     for(int i=;i<n;i++)
         printf("%d\n",b[i]);
 }