[洛谷P2396]yyy loves Maths VII $&$ [CF327E]Axis Walking

这道题是一个状压动归题。子集生成，每一位表示是否选择了第$i$个数。

转移：$f[S] = sum f[S-{x}]$且$xin S$，当该子集所有元素的和为$b_1$或$b_2$时不转移。

初始化：$f[{}]=1$，其他为$0$。

目标：$f[全集]$。

注意常数即可。

#include <bits/stdc++.h>

using namespace std;

#define re register
#define rep(i, a, b) for (re int i = a; i <= b; ++i)
#define repd(i, a, b) for (re int i = a; i >= b; --i)
#define maxx(a, b) a = max(a, b);
#define minn(a, b) a = min(a, b);
#define LL long long
#define inf (1 << 30)

inline int read() {
    int w = 0, f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

int f[1 << 24], dis[1 << 24], n, m, b[3];
const int Mod = 1000000007;

#define mod(x) (x>=Mod?x-Mod:x)

int main() {
    n = read();
    rep(i, 1, n) dis[1<<(i-1)] = read();
    m = read();
    rep(i, 1, m) b[i] = read();
    f[0] = 1;
    rep(i, 1, (1<<n)) {
        dis[i] = dis[i&(i-1)] + dis[i&-i];
        if (dis[i] == b[1] || dis[i] == b[2]) continue;
        for (register int x = i; x; x &= (x-1)) f[i] = mod(f[i]+f[i^x&-x]);
    }
    printf("%d", f[(1<<n)-1]);
    return 0;
}