BZOJ4589: Hard Nim(FWT 快速幂)

题意

题目链接

Sol

神仙题Orzzzz

题目可以转化为从\(\leqslant M\)的质数中选出\(N\)个\(xor\)和为\(0\)的方案数

这样就好做多了

设\(f(x) = [x \text{是质数}]\)

\(n\)次异或FWT即可

快速幂优化一下，中间不用IFWT，最后转一次就行(然而并不知道为什么)

哪位大佬教教我这题的DP怎么写呀qwqqqq

死过不过去样例。。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = (1 << 17) + 10, mod = 998244353, inv2 = 499122177;
inline int read() {
	char c = getchar(); int x = 0, f = 1;
	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
int N, A[MAXN], B[MAXN], C[MAXN];
int add(int x, int y) {
	if(x + y < 0) return x + y + mod;
	return x + y >= mod ? x + y - mod : x + y;
}
int mul(int x, int y) {
	return 1ll * x * y % mod;
}
void FWTor(int *a, int opt) {
	for(int mid = 1; mid < N; mid <<= 1) 
		for(int R = mid << 1, j = 0; j < N; j += R)
			for(int k = 0; k < mid; k++) 
				if(opt == 1) a[j + k + mid] = add(a[j + k], a[j + k + mid]);
				else a[j + k + mid] = add(a[j + k + mid], -a[j + k]);
}
void FWTand(int *a, int opt) {
	for(int mid = 1; mid < N; mid <<= 1) 
		for(int R = mid << 1, j = 0; j < N; j += R)
			for(int k = 0; k < mid; k++) 
				if(opt == 1) a[j + k] = add(a[j + k], a[j + k + mid]);
				else a[j + k] = add(a[j + k], -a[j + k + mid]);
}
void FWTxor(int *a, int opt) {
	for(int mid = 1; mid < N; mid <<= 1) 
		for(int R = mid << 1, j = 0; j < N; j += R)
			for(int k = 0; k < mid; k++) {
				int x = a[j + k], y = a[j + k + mid];
				if(opt == 1) a[j + k] = add(x, y), a[j + k + mid] = add(x, -y);
				else a[j + k] = mul(add(x, y), inv2), a[j + k + mid] = mul(add(x, -y), inv2); 				
			}

}
int main() {
	N = 1 << (read());
	for(int i = 0; i < N; i++) A[i] = read();
	for(int i = 0; i < N; i++) B[i] = read();
	FWTor(A, 1); FWTor(B, 1);
	for(int i = 0; i < N; i++) C[i] = mul(A[i], B[i]);
	FWTor(C, -1); FWTor(A, -1); FWTor(B, -1);
	for(int i = 0; i < N; i++) printf("%d ", C[i]); puts("");
	FWTand(A, 1); FWTand(B, 1);
	for(int i = 0; i < N; i++) C[i] = mul(A[i], B[i]);
	FWTand(C, -1); FWTand(A, -1); FWTand(B, -1);	
	for(int i = 0; i < N; i++) printf("%d ", C[i]); puts("");
	FWTxor(A, 1); FWTxor(B, 1);
	for(int i = 0; i < N; i++) C[i] = mul(A[i], B[i]);
	FWTxor(C, -1); FWTxor(A, -1); FWTxor(B, -1);	
	for(int i = 0; i < N; i++) printf("%d ", C[i]);
	return 0;
}