题意

给定n和 a[i] b[i],
已知FWT(a[i]) * FWT(ans[i]) = FWT(b[i]),求ans[i]
(XOR)

题解

FWT板题，赛后学习了一个

AC代码

#include <iostream>

using namespace std;
using ll = long long;
ll MOD = 1e9+7;
ll inv2;
int n;

const int maxn = (1 << 20) + 5;
ll a[maxn],b[maxn];

ll qpow(ll x, ll y) {
    ll re = 1;
    for (; y > 0; y >>= 1) {
        if (y & 1) {
            re = re * x % MOD;
        }
        x = x * x % MOD;
    }
    return re;
}

void FWT_xor(ll *a,int opt)
{
    for(int i=1;i<n;i<<=1)
        for(int p=i<<1,j=0;j<n;j+=p)
            for(int k=0;k<i;++k)
            {
                ll 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==-1)a[j+k]=1ll*a[j+k]*inv2%MOD,a[i+j+k]=1ll*a[i+j+k]*inv2%MOD;
            }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    inv2 = qpow(2,MOD-2);
    cin >> n;
    for(int i = 0;i<n;++i)cin >> a[i];
    for(int i = 0;i<n;++i)cin >> b[i];
    FWT_xor(a,1);
    FWT_xor(b,1);
    for(int i = 0;i<n;++i) b[i] = b[i] * qpow(a[i],MOD - 2) % MOD;
    FWT_xor(b,-1);
    for(int i = 0;i<n;++i) cout << b[i] << "\n";
}