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鏈接：https://www.nowcoder.com/acm/contest/147/A
來源：牛客網

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.

輸入描述:

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

輸出描述:

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.

示例1

輸入

4
1 10 100 1000
1234 2143 3412 4321

輸出4

3
2
1


解析 　　給出        已知A b 求 x 

上式可以轉化成           由於 i^j^j=i   


所以原式等式            直接套fwt_xor 板子
 

代碼

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define rep(i,a,n) for (int i=a;i<n;i++)
 4 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 5
 
 #define pb push_back
 6 #define mp make_pair
 7 #define all(x) (x).begin(),(x).end()
 8 #define fi first
 9 #define se second
10 #define SZ(x) ((int)(x).size())
11 typedef vector<int> VI;
12 typedef long long ll;
13 typedef pair<int,int> PII;
14 const ll mod=1000000007;
15 const int maxn = 3e6+10;
16 ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
17 // head
18 
19 int a[maxn],b[maxn];
20 void FWT_or(int *a,int N,int opt)
21 {
22     for(int i=1;i<N;i<<=1)
23         for(int p=i<<1,j=0;j<N;j+=p)
24             for(int k=0;k<i;++k)
25                 if(opt==1)a[i+j+k]=(a[j+k]+a[i+j+k])%mod;
26                 else a[i+j+k]=(a[i+j+k]+mod-a[j+k])%mod;
27 }
28 void FWT_and(int *a,int N,int opt)
29 {
30     for(int i=1;i<N;i<<=1)
31         for(int p=i<<1,j=0;j<N;j+=p)
32             for(int k=0;k<i;++k)
33                 if(opt==1)a[j+k]=(a[j+k]+a[i+j+k])%mod;
34                 else a[j+k]=(a[j+k]+mod-a[i+j+k])%mod;
35 }
36 void FWT_xor(int *a,int N,int opt) //opt=1 正變換 opt=-1 逆變換 
37 {
38     ll inv2=powmod(2,mod-2);
39     for(int i=1;i<N;i<<=1)
40         for(int p=i<<1,j=0;j<N;j+=p)
41             for(int k=0;k<i;++k)
42             {
43                 int X=a[j+k],Y=a[i+j+k];
44                 a[j+k]=(X+Y)%mod;a[i+j+k]=(X+mod-Y)%mod;
45                 if(opt==-1)a[j+k]=1ll*a[j+k]*inv2%mod,a[i+j+k]=1ll*a[i+j+k]*inv2%mod;
46             }
47 }
48 int main()
49 {
50     int n;
51     scanf("%d",&n);
52     for(int i=0;i<n;i++)
53         scanf("%d",&a[i]);
54     for(int i=0;i<n;i++)
55         scanf("%d",&b[i]);
56     FWT_xor(a,n,1);FWT_xor(b,n,1);
57     for(int i=0;i<n;i++)
58     {
59         b[i]=b[i]*powmod(a[i],mod-2)%mod;   //  b/a 
60     }
61     FWT_xor(b,n,-1);   //再逆變換
62     for(int i=0;i<n;i++)
63         printf("%d\n",b[i]);
64 }

牛客網暑期ACM多校訓練營（第九場） A題 FWT