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

E. Anya and Cubes

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 kstickers 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 kexclamation 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.

Sample test(s)

input

2 2 30
4 3

output

1

input

2 2 7
4 3

output

1

input

3 1 11 1 1

output

6

Note

In the first sample the only way is to choose both cubes and stick an exclamation mark on each of them.

In the second sample the only way is to choose both cubes but don't stick an exclamation mark on any of them.

In the third sample it is possible to choose any of the cubes in three ways, and also we may choose to stick or not to stick the exclamation mark on it. So, the total number of ways is six.

参考链接： http://www.cnblogs.com/qscqesze/p/4371851.html
给你n个数，k个魔法棒，s为所求的数，然后让你找有多少种方法，能够使的这n个数之和为s，其中一个魔法棒可以使的一个数变成他的阶乘

折半搜索
对于每一个数，都有三种策略，拿，不拿，使用魔法棒
那么我们就直接折半搜索然后扔进一个map里面，然后就查询

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef long long ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 30

ll a
,n,k,s;
ll mid;

struct stud{
ll step,result;
}f[1111111];

ll p[20];
ll cnt;

map<ll,ll>mp[30];

void inint()
{
	int i,j;
	p[0]=1;
	p[1]=1;
	fre(i,2,20)
	 p[i]=p[i-1]*i;

}

void dfs(ll pos,ll num,ll va)
{
     if(num>k||va>s) return ;
	
		
     if(pos==mid+1)
	 {
	 	f[cnt].step=num;
	 	f[cnt++].result=va;
	 	return ;
	 }

    dfs(pos+1,num,va);
    dfs(pos+1,num,va+a[pos]);
    if(a[pos]<20)
		dfs(pos+1,num+1,va+p[a[pos]]);

}

void DFS(ll pos,ll num,ll va)
{
   if(num>k||va>s) return ;

   if(pos==n+1)
	{
	   mp[num][va]++;
	   return ;
	}

   DFS(pos+1,num,va);
    DFS(pos+1,num,va+a[pos]);
    if(a[pos]<20)
		DFS(pos+1,num+1,va+p[a[pos]]);

}

int main()
{
	ll i,j;
	inint();

	while(~scanf("%I64d%I64d%I64d",&n,&k,&s))
	{
		fre(i,1,n+1)
		  scanf("%I64d",&a[i]);

        fre(i,0,30)
		  mp[i].clear();

		mid=MID(0,n);
		cnt=0;

		dfs(1,0,0);

		DFS(mid+1,0,0);
        ll ans=0;

        fre(j,0,cnt)
          fre(i,0,k-f[j].step+1)
            ans+=mp[i][s-f[j].result];

    pf("%I64d\n",ans);
	}
	return 0;
}