Barbells

Your local gym has n barbells and m plates. In order to prepare a weight for lifting, you mustchoose a single barbell, which has two sides. You then load each side with a (possibly empty) setof plates. For safety reasons, the plates on each side must sum to the same weight. What weightsare available for lifting?

For example, suppose that there are two barbells weighing $100$ and $110$ grams, and five platesweighting $1, 4, 5, 5,$ and $6$ grams, respectively. Then, there are six possible weights available forlifting. The table below shows one way to attain the different weights:

Barbell	Left side	Right side	Total
100	0	0	100
100	5	5	110
100	1 + 5	6	112
110	5	5	120
110	1 + 5	6	122
110	5 + 5	4 + 6	130

Input

The first line of input contains the space-separated integers $n$ and $m$ $(1 ≤ n, m ≤ 14)$.The second line of input contains $n$ space-separated integers $b_1, ..., b_n$ $(1 ≤ b_i ≤ 10^8)$, denoting theweights of the barbells in grams.The third line of input contains m space-separated integers $p_1, ..., p_m \ (1 ≤ p_i ≤ 10^8)$, denoting theweights of the plates in grams.

Output

Print a sorted list of all possible distinct weights in grams, one per line.

樣例輸入
2 5
100 110
5 5 1 4 6
樣例輸出
100
110
112
120
122
130

$n$和$m$的數量較少，最多隻有$14$個，因此暴搜＋剪枝即可。

#include<bits/stdc++.h>
#include<string.h>

#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define sc(a) scahf("%c",&a);
#define ss(a) scanf("%s",a)
#define pi(a) printf("%d\n",a)
#define pl(a) printf("%lld\n",a)
#define pc(a) putchar(a)
#define ms(a) memset(a,0,sizeof(a))
#define repi(i, a, b) for(register int i=a;i<=b;++i)
#define repd(i, a, b) for(register int i=a;i>=b;--i)
#define reps(s) for(register int i=head[s];i;i=Next[i])
#define ll long long
#define ull unsigned long long
#define vi vector<int>
#define pii pair<int,int>
#define mii unordered_map<int,int>
#define msi unordered_map<string,int>
#define lowbit(x) ((x)&(-(x)))
#define ce(i, r) i==r?'\n':' '
#define pb push_back
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define pr(x) cout<<#x<<": "<<x<<endl
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 20;
bool v[N];
vector<ll> ans;
int a[N], b[N], n, m;
ll sum;

void dfs2(int u, ll sum1, ll sum2) {
    if (u == n + 1) {
        if (sum1 == sum2)repi(i, 1, m)ans.pb(sum1 * 2 + b[i]);
        return;
    }
    if (sum2 > sum1)return;
    if (!v[u]) {
        v[u] = true;
        dfs2(u + 1, sum1, sum2 + a[u]);
        v[u] = false;
    }
    dfs2(u + 1, sum1, sum2);
}

void dfs1(int u, ll sum1) {
    if (sum1 > sum)return;
    if (u == n + 1) {
        dfs2(1, sum1, 0);
        return;
    }
    v[u] = true;
    dfs1(u + 1, sum1 + a[u]);
    v[u] = false;
    dfs1(u + 1, sum1);
}

int main() {
    m = qr(), n = qr();
    repi(i, 1, m)b[i] = qr();
    repi(i, 1, n)a[i] = qr(), sum += a[i];
    sum >>= 1;
    dfs1(1, 0);
    sort(ans.begin(), ans.end());
    ans.erase(unique(ans.begin(), ans.end()), ans.end());
    for (auto it:ans)pl(it);
    return 0;
}