Codeforces Round #191 (Div. 二) E. Axis Walking (状态压缩+lowbit应用)

Codeforces Round #191 (Div. 2) E. Axis Walking (状态压缩+lowbit应用)

Iahub wants to meet his girlfriend Iahubina. They both live in Ox axis (the horizontal axis). Iahub lives at point 0 and Iahubina at point d.

Iahub has n positive integers a1, a2, ..., an. The sum of those numbers is d. Suppose p1, p2, ..., pn is a permutation of {1, 2, ..., n}. Then, let b1 = ap1, b2 = ap2 and so on. The array b is called a "route". There are n! different routes, one for each permutation p.

Iahub's travel schedule is: he walks b1 steps on Ox axis, then he makes a break in point b1. Then, he walks b2 more steps on Ox axis and makes a break in point b1 + b2. Similarly, at j-th (1 ≤ j ≤ n) time he walks bj more steps on Ox axis and makes a break in pointb1 + b2 + ... + bj.

Iahub is very superstitious and has k integers which give him bad luck. He calls a route "good" if he never makes a break in a point corresponding to one of those k numbers. For his own curiosity, answer how many good routes he can make, modulo 1000000007(109 + 7).

Input

The first line contains an integer n (1 ≤ n ≤ 24). The following line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 109).

The third line contains integer k (0 ≤ k ≤ 2). The fourth line contains k positive integers, representing the numbers that give Iahub bad luck. Each of these numbers does not exceed 109.

Output

Output a single integer — the answer of Iahub's dilemma modulo 1000000007 (109 + 7).

Sample test(s)

input
3
2 3 5
2
5 7

output
1

input
3
2 2 2
2
1 3

output
6

Note

In the first case consider six possible orderings:

[2, 3, 5]. Iahub will stop at position 2, 5 and 10. Among them, 5 is bad luck for him.
[2, 5, 3]. Iahub will stop at position 2, 7 and 10. Among them, 7 is bad luck for him.
[3, 2, 5]. He will stop at the unlucky 5.
[3, 5, 2]. This is a valid ordering.
[5, 2, 3]. He got unlucky twice (5 and 7).
[5, 3, 2]. Iahub would reject, as it sends him to position 5.

In the second case, note that it is possible that two different ways have the identical set of stopping. In fact, all six possible ways have the same stops: [2, 4, 6], so there's no bad luck for Iahub.

大致题意：

24项的数列，求有多少种排列使a1+a2+...+an的过程中不会产生b1和b2

大致思路：

状态压缩，每个状态表示这些项有多少种排列不会产生b1和b2

所以有：

for(int i = 0; (1<<i) <= mask; i++){
if( ((1<<i)&mask) == 0) continue;
dp[mask] += dp[ mask&(~(1<<i)) ];
}

复杂度是：

(1<<24)*24,复杂度太高

用lowbit优化，就不需要把i逐项移动找1了

复杂度接近(1<<24)*24/2

会优化一半，新技能get

////GNU C++11	Accepted	1278 ms	197000 KB

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
using namespace std;
typedef long long ll;
#define X first
#define Y second
typedef pair<int,int> pii;

template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x < 0) {
		putchar('-');
		x = -x;
	}
	if (x > 9) pt(x / 10);
	putchar(x % 10 + '0');
}


const int MOD = 1e9+7;
const int N = 30;
int b[N];
int dp[(1<<24)+5];
ll sum[(1<<24)+5];

inline int lowbit(int x){ return x&(-x);}
int main(){

        b[1] = b[2] = -1;
        int n,m;
        rd(n);
        for(int i = 0; i < n; i++) rd(sum[1<<i]);
        rd(m);
        REP(i,m) rd(b[i]);
        int all = (1<<n)-1;
        dp[0] = 1;
        REP(mask,all){
                if(sum[mask] == 0) sum[mask] = sum[mask-lowbit(mask)] + sum[lowbit(mask)];
                if(sum[mask] == b[1] || sum[mask] == b[2]) continue;
                ll t = 0;
                for( int tmp = mask; tmp ;tmp -= lowbit(tmp))
                        t += dp[mask-lowbit(tmp)];
                dp[mask] = t%MOD;
        }
        printf("%d\n",(int)dp[all]);
}

Codeforces Round #191 (Div. 二) E. Axis Walking (状态压缩+lowbit应用)

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