CF914G Sum the Fibonacci （快速沃尔什变换FWT + 子集卷积）

题面

题解

这是一道FWT和子集卷积的应用题。

我们先设 cnt[x] 表示 Si = x 的 i 的数量，那么

这里的Nab[x]指满足条件的 Sa|Sb=x、Sa&Sb=0 的(a,b)二元组数量，这个可以通过子集卷积快速求出，复杂度为

然后又设

那么就把答案简化为了

我们可以再次简化，设

这里的Nde[x]指满足条件的 Sd^Se=x 的(d,e)二元组数量，用FWT卷积求出，那么如果

就可以把答案简化为

最后考虑枚举  ，设答案为

所以我们就把它转化为了卷积的形式，用FWT这道题就完了。

ＣＯＤＥ

tym要AK了 %%%

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
#include<algorithm>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#define MAXN (1<<17|5)
#define LL long long
#define lowbit(x) (-(x) & (x))
#define ENDL putchar('\n')
#define rg register
//#pragma GCC optimize(2)
//#pragma G++ optimize(3)
//#define int LL
char char_read_before = 1;
inline int read() {
	int f = 1,x = 0;char s = char_read_before;
	while(s < '0' || s > '9') {if(s == '-') f = -1;s = getchar();}
	while(s >= '0' && s <= '9') {x = x * 10 - '0' + s;s = getchar();}
	char_read_before = s;return x * f;
}
inline int readN() {
	int x = 0;char s = char_read_before;
	while(s < '0' || s > '9') s = getchar();
	while(s >= '0' && s <= '9') {x = ((x<<3)+(x<<1)) + (s ^ 48);s = getchar();}
	char_read_before = s;return x;
}
inline int readone() {
	int x = 0;char s = getchar();
	while(s < '0' || s > '9') s = getchar();
	char_read_before = 1;return s - '0';
}
int zxy = 1000000007; // 用来膜的
int inv2 = (zxy+1)/2;
inline int qm(LL x,int dalao) {return x >= dalao ? qm(x-dalao,dalao):x;}
int n,m,i,j,s,o,k;
inline void DWTXOR(int *s,int m) {
	for(int k = m;k > 1;k >>= 1) {
		for(int i = 0;i < m;i += k) {
			for(int j = i+(k>>1);j < i+k;j ++) {
				int s0 = s[j-(k>>1)],s1 = s[j];
				s[j] = qm((s0 +0ll+ zxy - s1) , zxy);
				s[j-(k>>1)] = qm((s0 +0ll+ s1) , zxy);
			}
		}
	}
	return ;
}
inline void IDWTXOR(int *s,int m) {
	for(int k = 2;k <= m;k <<= 1) {
		for(int i = 0;i < m;i += k) {
			for(int j = i+(k>>1);j < i+k;j ++) {
				int s0 = s[j-(k>>1)],s1 = s[j];
				s[j-(k>>1)] = qm((s0 +0ll+ s1) , zxy) *1ll* inv2 % zxy;
				s[j] = qm((s0 +0ll+ zxy - s1) , zxy) *1ll* inv2 % zxy;
			}
		}
	}
	return ;
}
inline void DWTOR(int *s,int m) {
	for(int k = m;k > 1;k >>= 1) {
		for(int i = 0;i < m;i += k) {
			for(int j = i+(k>>1);j < i+k;j ++) {
				int s0 = s[j-(k>>1)],s1 = s[j];
				s[j] = qm((s0 +0ll+ s1) , zxy);
			}
		}
	}
	return ;
}
inline void IDWTOR(int *s,int m) {
	for(int k = 2;k <= m;k <<= 1) {
		for(int i = 0;i < m;i += k) {
			for(int j = i+(k>>1);j < i+k;j ++) {
				int s0 = s[j-(k>>1)],s1 = s[j];
				s[j] = qm((s1 +0ll+ zxy - s0) , zxy);
			}
		}
	}
	return ;
}
inline void DWTAND(int *s,int m) {
	for(int k = m;k > 1;k >>= 1) {
		for(int i = 0;i < m;i += k) {
			for(int j = i+(k>>1);j < i+k;j ++) {
				LL s0 = s[j-(k>>1)],s1 = s[j];
				s[j-(k>>1)] = qm((s0 +0ll+ s1) , zxy);
			}
		}
	}
	return ;
}
inline void IDWTAND(int *s,int m) {
	for(int k = 2;k <= m;k <<= 1) {
		for(int i = 0;i < m;i += k) {
			for(int j = i+(k>>1);j < i+k;j ++) {
				int s0 = s[j-(k>>1)],s1 = s[j];
				s[j-(k>>1)] = qm((s0 +0ll+ zxy - s1) , zxy);
			}
		}
	}
	return ;
}
int fb[MAXN];
int A[23][MAXN],B[23][MAXN],AB[23][MAXN];
int ab[MAXN];
int C[MAXN],D[MAXN],E[MAXN],DE[MAXN];
int ct[MAXN],as[MAXN];
int main() {
	n = read();
	fb[1] = 1;
	for(int i = 1;i < (1<<17);i ++) {
		ct[i] = ct[i^lowbit(i)] + 1;
		if(i-1) fb[i] = qm(fb[i-2] +0ll+ fb[i-1],zxy);
	}
	int maxn = 0;
	for(int i = 1;i <= n;i ++) {
		s = read();
		maxn = max(maxn,s);
		A[ct[s]][s] ++;
		B[ct[s]][s] ++;
		C[s] ++;D[s] ++;E[s]++;
	}
	n = 0;
	m = 1;while(m <= maxn) m <<= 1,n++;
	for(int i = 0;i <= n;i ++) {
		DWTOR(A[i],m);
		DWTOR(B[i],m);
	}
	DWTXOR(D,m);DWTXOR(E,m);
	for(int i = 0;i < m;i ++) DE[i] = D[i] *1ll* E[i] % zxy;
	IDWTXOR(DE,m);
	for(int i = 0;i <= n;i ++) {
		for(int j = 0;j <= i;j ++) {
			for(int k = 0;k < m;k ++) {
				AB[i][k] = qm((AB[i][k] +0ll+ A[j][k] *1ll* B[i-j][k] % zxy),zxy);
			}
		}
		IDWTOR(AB[i],m);
	}
	for(int i = 0;i < m;i ++) {
		ab[i] = AB[ct[i]][i];
		ab[i] = ab[i] *1ll* fb[i] % zxy;
		C[i] = C[i] *1ll* fb[i] % zxy;
		DE[i] = DE[i] *1ll* fb[i] % zxy;
	}
	DWTAND(ab,m);
	DWTAND(C,m);
	DWTAND(DE,m);
	for(int i = 0;i < m;i ++) {
		as[i] = ab[i] *1ll* C[i] % zxy *1ll* DE[i] % zxy;
	}
	IDWTAND(as,m);
	int ans = 0;
	for(int i = 0;i <= n;i ++) {
		ans = qm(ans +0ll+ as[1<<i],zxy);
	}
	printf("%d\n",ans);
	return 0;
}