题目大意

给出\(n,s_{1,2,...,n}\)，定义一个五元组\((a,b,c,d,e)\)合法当且仅当:

\[1\le a,b,c,d,e\le n
\]
\[(s_a\vee s_b)\wedge s_c \wedge (s_d\oplus s_e)=2^i,i\in \mathbb{Z}
\]
\[s_a\wedge s_b=0
\]

求出对于所有合法的五元组\((a,b,c,d,e)\):

\[\sum f(s_a\vee s_b)f(s_c)f(s_d\oplus s_e)
\]

其中\(f(i)\)表示第\(i\)位斐波拉契数列。

思路

其实这个题应该算\(\text {FST}\)的入门级题目，也不是很难。

首先，我们定义\(v(a,b,c,d,e)=(s_a\vee s_b)\wedge s_c \wedge (s_d\oplus s_e)\)。于是我们可以把式子写成这样一个形式:

\[\sum_{i} \sum_{v(a,b,c,d,e)=2^i} [s_a\wedge s_b=0]f(s_a\vee s_b)f(s_c)f(s_d\oplus s_e)
\]

\[=\sum_{p} \sum_{i\wedge j\wedge k=2^p} f(i)f(j)f(k)(\sum_{s_a\vee s_b=i,s_a\wedge s_b=0}1)(\sum_{s_a\oplus s_b=k}1)
\]

然后我们就发现第一个括号里面的可以用子集卷积求到，后面那个可以用异或卷积求到，总的又可以用并卷积求到。于是我们就可以在\(\Theta(w\log ^2w)\)的时间内求到了。其中\(w\)是值域。

\(\text {Code}\)

#include <bits/stdc++.h>

using namespace std;

#define Int register int

#define inv2 500000004

#define mod 1000000007

#define MAXN 1000005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}

template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}

template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

int lim = 1; 

void mul (int &a,int b){a = 1ll * a * b % mod;}

void del (int &a,int b){a = a >= b ? a - b : a + mod - b;}

void add (int &a,int b){a = a + b >= mod ? a + b - mod : a + b;}

void ORFWT (int *A,int type){

	for (Int i = 1;i < lim;i <<= 1)

		for (Int j = 0;j < lim;j += i << 1)

			for (Int k = 0;k < i;++ k)

				if (type == 1) add (A[i + j + k],A[j + k]);

				else del (A[i + j + k],A[j + k]);

}

void ANDFWT (int *A,int type){

	for (Int i = 1;i < lim;i <<= 1)

		for (Int j = 0;j < lim;j += i << 1)

			for (Int k = 0;k < i;++ k)

				if (type == 1) add (A[j + k],A[i + j + k]);

				else del (A[j + k],A[i + j + k]);

}

void XORFWT (int *A,int type){

	for (Int i = 1;i < lim;i <<= 1)

		for (Int j = 0;j < lim;j += i << 1)

			for (Int k = 0;k < i;++ k){

				int x = A[j + k],y = A[i + j + k];

				if (type == 1) A[j + k] = (x + y) % mod,A[i + j + k] = (x + mod - y) % mod;

				else A[j + k] = 1ll * (x + y) * inv2 % mod,A[i + j + k] = 1ll * (x + mod - y) * inv2 % mod;

			}

}

int n,s,fib[1 << 17],cnt[1 << 17],A[1 << 17],S[1 << 17],f[18][1 << 17],h[1 << 17],sum[1 << 17];

signed main(){

	read (n);

	fib[0] = 0,fib[1] = cnt[1] = 1;int maxn = 0;

	for (Int i = 2;i < (1 << 17);++ i) fib[i] = (fib[i - 1] + fib[i - 2]) % mod,cnt[i] = cnt[i >> 1] + (i & 1);

	for (Int i = 1,s;i <= n;++ i) read (s),maxn = max (maxn,s),add (f[cnt[s]][s],1),add (h[s],1),add (sum[s],fib[s]);

	int logn = 0;while (lim <= maxn) lim <<= 1,logn ++;

	for (Int i = 0;i <= logn;++ i) ORFWT (f[i],1);

	for (Int i = 0;i <= logn;++ i){

		for (Int j = 0;j < lim;++ j) S[j] = 0;

		for (Int j = 0;j <= i;++ j)

			for (Int k = 0;k < lim;++ k)

				add (S[k],1ll * f[j][k] * f[i - j][k] % mod);

		ORFWT (S,-1);

		for (Int j = 0;j < lim;++ j) if (cnt[j] == i) add (A[j],S[j]);

	}

	XORFWT (h,1);

	for (Int i = 0;i < lim;++ i) mul (h[i],h[i]);

	XORFWT (h,-1);

	for (Int i = 0;i < lim;++ i) mul (A[i],fib[i]),mul (h[i],fib[i]);

	ANDFWT (A,1),ANDFWT (h,1),ANDFWT (sum,1);

	for (Int i = 0;i < lim;++ i) mul (A[i],h[i]),mul (A[i],sum[i]);

	ANDFWT (A,-1);

	int ans = 0;for (Int i = 1;i < lim;i <<= 1) add (ans,A[i]);

	write (ans),putchar ('\n');

	return 0;

}

巴特西

题解 CF914G Sum the Fibonacci

题目大意

思路

\(\text {Code}\)

最新文章

热门文章