[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$x\in S$，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 = , f = ; char c = getchar();
     while (!isdigit(c)) f = c == '-' ? - : f, c = getchar();
     while (isdigit(c)) w = (w << ) + (w << ) + (c ^ ''), c = getchar();
     return w * f;
 }

 int f[ << ], dis[ << ], n, m, b[];
 const int Mod = ;

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

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