题解
FWT板子题
可以发现
(dp[i][u] = sum_{i = 0}^{N - 1} dp[i - 1][u xor (2^i)] + dp[i - 1][u])
然后如果把异或提出来可以变成一个异或卷积
也就是另一个数组里只有(0),(2^0),(2^1)...(2^{n - 1})有值
用FWT变换一下,然后快速幂,之后和原数组卷积起来就是答案了
代码
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define enter putchar('
')
#define space putchar(' ')
#define MAXN 2000005
//#define ivorysi
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 - '0' + c;
c = getchar();
}
res = res * f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,T,g[MAXN],f[MAXN],Inv2 = (MOD + 1) / 2;
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
void FWT(int *a) {
for(int i = 1 ; i < (1 << N) ; i <<= 1) {
for(int j = 0 ; j < (1 << N) ; j += (i << 1)) {
for(int k = 0 ; k < i ; ++k) {
int t0 = a[j + k],t1 = a[j + k + i];
a[j + k] = inc(t0,t1);
a[j + k + i] = inc(t0,MOD - t1);
}
}
}
}
void IFWT(int *a) {
for(int i = 1 ; i < (1 << N) ; i <<= 1) {
for(int j = 0 ; j < (1 << N) ; j += (i << 1)) {
for(int k = 0 ; k < i ; ++k) {
int t0 = a[j + k],t1 = a[j + k + i];
a[j + k] = mul(inc(t0,t1),Inv2);
a[j + k + i] = mul(inc(t0,MOD - t1),Inv2);
}
}
}
}
void conv(int *a,int *b) {
for(int i = 0 ; i < (1 << N) ; ++i) a[i] = mul(a[i],b[i]);
}
void fpow(int *a,int *ans,int c) {
static int t[MAXN];
for(int i = 0 ; i < (1 << N) ; ++i) t[i] = a[i],ans[i] = a[i];
--c;
while(c) {
if(c & 1) conv(ans,t);
conv(t,t);
c >>= 1;
}
}
void Solve() {
read(N);read(T);
g[0] = 1;
for(int i = 0 ; i < N ; ++i) g[1 << i] = 1;
for(int i = 0 ; i < (1 << N) ; ++i) read(f[i]);
FWT(g);FWT(f);
fpow(g,g,T);
conv(f,g);
IFWT(f);
for(int i = 0 ; i < (1 << N) ; ++i) {
out(f[i]);if(i != (1 << N) - 1) space;
}
enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}