[SOJ #47]集合并卷积

题目大意：给你两个多项式$A,B$，求多项式$C$使得：
$$
C_n=\sum\limits_{x|y=n}A_xB_y
$$
题解：$FWT$，他可以解决形如$C_n=\sum\limits_{x\oplus y=n}A_xB_y$的问题，其中$\oplus$为位运算（一般为$or,and,xor$）

or：

void FWT(int *A) {
	for (int mid = 1; mid < lim; mid <<= 1)
		for (int i = 0; i < lim; i += mid << 1)
			for (int j = 0; j < mid; ++j) A[i + j + mid] += A[i + j];
}
void IFWT(int *A) {
	for (int mid = 1; mid < lim; mid <<= 1)
		for (int i = 0; i < lim; i += mid << 1)
			for (int j = 0; j < mid; ++j) A[i + j + mid] -= A[i + j];
}

　　

and：

void FWT(int *A) {
	for (int mid = 1; mid < lim; mid <<= 1)
		for (int i = 0; i < lim; i += mid << 1)
			for (int j = 0; j < mid; ++j) {
				int X = A[i + j], Y = A[i + j + mid];
				A[i + j] = X + Y, A[i + j + mid] = X - Y;
			}
}
void IFWT(int *A) {
	for (int mid = 1; mid < lim; mid <<= 1)
		for (int i = 0; i < lim; i += mid << 1)
			for (int j = 0; j < mid; ++j) {
				int X = A[i + j], Y = A[i + j + mid];
				A[i + j] = X + Y, A[i + j + mid] = X - Y;
			}
	for (int i = 0; i < lim; ++i) A[i] /= lim;
}

　　

xor：

void FWT(int *A) {
	for (int mid = 1; mid < lim; mid <<= 1)
		for (int i = 0; i < lim; i += mid << 1)
			for (int j = 0; j < mid; ++j) {
				int X = A[i + j], Y = A[i + j + mid];
				A[i + j] = X + Y, A[i + j + mid] = X - Y;
			}
}
void IFWT(int *A) {
	for (int mid = 1; mid < lim; mid <<= 1)
		for (int i = 0; i < lim; i += mid << 1)
			for (int j = 0; j < mid; ++j) {
				int X = A[i + j], Y = A[i + j + mid];
				A[i + j] = X + Y, A[i + j + mid] = X - Y;
			}
	for (int i = 0; i < lim; ++i) A[i] /= lim;
}

　　

卡点：无

 

C++ Code：

#include 
#include 
inline int read() {
	static int ch;
	while (isspace(ch = getchar())) ;
	return ch & 15;
}

#define N 1048576
int lim;
inline void init(const int n) {
	lim = 1; while (lim < n) lim <<= 1;
}
inline void FWT(long long *A) {
	for (int mid = 1; mid < lim; mid <<= 1) 
		for (int i = 0; i < lim; i += mid << 1) 
			for (int j = 0; j < mid; ++j) A[i + j + mid] += A[i + j];
}
inline void IFWT(long long *A) {
	for (int mid = 1; mid < lim; mid <<= 1) 
		for (int i = 0; i < lim; i += mid << 1) 
			for (int j = 0; j < mid; ++j) A[i + j + mid] -= A[i + j];
}

int n;
long long A[N], B[N];
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; ++i) A[i] = read();
	for (int i = 0; i < n; ++i) B[i] = read();
	init(n);
	FWT(A), FWT(B);
	for (int i = 0; i < lim; ++i) A[i] *= B[i];
	IFWT(A);
	for (int i = 0; i < n; ++i) {
		printf("%lld", A[i]);
		putchar(i == (n - 1) ? '\n' : ' ');
	}
	return 0;
}

　　

转载于:https://www.cnblogs.com/Memory-of-winter/p/10259231.html