【Codeforces Round #297 (Div. 2)】Codeforces 525E Anya and Cubes

Anya loves to fold and stick. Today she decided to do just that.

Anya has n cubes lying in a line and numbered from 1 to n from left to
right, with natural numbers written on them. She also has k stickers
with exclamation marks. We know that the number of stickers does not
exceed the number of cubes.

Anya can stick an exclamation mark on the cube and get the factorial
of the number written on the cube. For example, if a cube reads 5,
then after the sticking it reads 5!, which equals 120.

You need to help Anya count how many ways there are to choose some of
the cubes and stick on some of the chosen cubes at most k exclamation
marks so that the sum of the numbers written on the chosen cubes after
the sticking becomes equal to S. Anya can stick at most one
exclamation mark on each cube. Can you do it?

Two ways are considered the same if they have the same set of chosen
cubes and the same set of cubes with exclamation marks. Input

The first line of the input contains three space-separated integers n,
k and S (1 ≤ n ≤ 25, 0 ≤ k ≤ n, 1 ≤ S ≤ 1016) — the number of cubes
and the number of stickers that Anya has, and the sum that she needs
to get.

The second line contains n positive integers ai (1 ≤ ai ≤ 109) — the
numbers, written on the cubes. The cubes in the input are described in
the order from left to right, starting from the first one.

Multiple cubes can contain the same numbers. Output

Output the number of ways to choose some number of cubes and stick
exclamation marks on some of them so that the sum of the numbers
became equal to the given number S.

每个物品有三种决策，直接暴力枚举O(3^n)会超时，但是我们发现O(3^(n/2))不会超时，于是可以采用中途相遇法。
先暴力枚举一半的决策，把对应的值、使用次数和方案数存到map里，然后再枚举另一半，在map中查找。
复杂度O(3^(n/2)log(3^(n/2)))=O(3^(n/2)n)。

#include
#include
#include
#include
using namespace std;
#define LL long long
mapint> m[30];
int n,k;
LL s,a[30],fact[20],ans;
void dfs1(LL sum,int p,int now)
{
    if (p==n/2+1)
    {
        if (m[now].count(sum)) m[now][sum]++;
        else m[now][sum]=1;
        return;
    }
    dfs1(sum,p+1,now);
    if (a[p]<=s-sum) dfs1(sum+a[p],p+1,now);
    if (now19&&fact[a[p]]<=s-sum) dfs1(sum+fact[a[p]],p+1,now+1);
}
void dfs2(LL sum,int p,int now)
{
    if (p==n+1)
    {
        for (int i=0;i+now<=k;i++)
          if (m[i].count(s-sum))
            ans+=m[i][s-sum];
        return;
    }
    dfs2(sum,p+1,now);
    if (a[p]<=s-sum) dfs2(sum+a[p],p+1,now);
    if (now19&&fact[a[p]]<=s-sum) dfs2(sum+fact[a[p]],p+1,now+1);
}
int main()
{
    int i,j;
    LL x,y,z;
    fact[0]=1;
    for (i=1;i<=19;i++)
      fact[i]=fact[i-1]*i;
    scanf("%d%d%I64d",&n,&k,&s);
    for (i=1;i<=n;i++)
      scanf("%I64d",&a[i]);
    dfs1(0,1,0);
    dfs2(0,n/2+1,0);
    printf("%I64d\n",ans);
}