SP2829 TLE (FWT)

高维前缀和

众所周知, FWT可以轻松的算出高维前缀和

本题题解:

考虑状压\(dp\)(题目都说了\(2^M\)那就状压了)因为\(\%c[i]\)和\(\&a[i-1]\)这两个操作都和具体的数值有关

\(F[i][j]\)表示枚举到\(i\), 第\(i\)个数填\(j\)有多少种方案

\[F[i][j] = \begin{cases} 0~~~~~~~~~~~ j~\%~c[i] ==0\\\\\displaystyle \sum_{k\&j=0} F[i-1][j]\end{cases} \]

这样显然是不行的, 需要一点优化

转移时发现每一个合法的k都是\(j \bigotimes (2^M-1)\)的子集, 如果能记个子集前缀和那就再好不过了, FWT可以帮我们完成这件事, 一遍FWT_or算出子集和, 给出一份易于背诵的代码吧 戳这里

完整代码

#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;
const int P = 1000000000;
template <typename T>
void read(T &x) {
    x = 0; bool f = 0;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
    for (;isdigit(c);c=getchar()) x=x*10+(c^48);
    if (f) x=-x;
}

ll ans = 0, n, m;
void FWT_or(ll *f) {
	for (int i = 1, p = 2; i < m; i <<= 1, p <<= 1) 
		for (int j = 0; j < m; j += p) 
			for (int k = 0; k < i; k++) 
				(f[i+j+k] += f[j+k]) %= P;
}

const int N = 55;
ll f[N][(1<<15) + 5], a[N];

int main() {
	int T; read(T);
	while (T--) {
		memset(f, 0, sizeof(f));
		read(n), read(m); m = 1 << m;
		for (int i = 1; i <= n; i++) read(a[i]);
		for (int i = 1; i < m; i++) 
			if (i % a[1]) f[1][i] = 1;
		FWT_or(f[1]);
		for (int i = 2; i <= n; i++) {
			for (int j = 1; j < m; j++) {
				if (j % a[i] == 0) continue;
				f[i][j] += f[i-1][(m-1)^j];
			}
			FWT_or(f[i]);
		}
		cout << f[n][m-1] << endl;
	}
	return 0;
}