Anya and Cubes 搜索+map映射-白红宇

Anya and Cubes 搜索+map映射

阅读量：6331 次

发布时间：2019-06-22

本文共 4591 字，大约阅读时间需要 15 分钟。

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 a_i (1 ≤ a_i ≤ 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.

Examples

Input

2 2 30 4 3

Output

Input

2 2 7 4 3

Output

Input

3 1 1 1 1 1

Output

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.

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             #include
              
               #include
               
                #include
                
                 #include
                 
                  #include
                  
                   #include
                   
                    #include
                    
                     #include
                     
                      #include
                      
                       #include
                       
                        //#include
                        
                         //#pragma GCC optimize(2)using namespace std;#define maxn 200005#define inf 0x7fffffff//#define INF 1e18#define rdint(x) scanf("%d",&x)#define rdllt(x) scanf("%lld",&x)#define rdult(x) scanf("%lu",&x)#define rdlf(x) scanf("%lf",&x)#define rdstr(x) scanf("%s",x)#define mclr(x,a) memset((x),a,sizeof(x))typedef long long ll;typedef unsigned long long ull;typedef unsigned int U;#define ms(x) memset((x),0,sizeof(x))const long long int mod = 1e9 + 7;#define Mod 1000000000#define sq(x) (x)*(x)#define eps 1e-5typedef pair
                         
                           pii;#define pi acos(-1.0)//const int N = 1005;#define REP(i,n) for(int i=0;i<(n);i++)typedef pair
                          
                            pii;inline int rd() { int x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x;}ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b);}int sqr(int x) { return x * x; }/*ll ans;ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans;}*/int n, K;int val[maxn];ll sum;ll ans;typedef pair
                           
                            pli;map
                            
                             a, b;ll fac[25];void dfs1(ll res, int pos, int k) { if (res > sum)return; if (k > K)return; if (pos > n / 2) { a[pli(res, k)]++; return; } dfs1(res, pos + 1, k); dfs1(res + val[pos], pos + 1, k); if (val[pos] <= 20) { dfs1(res + fac[val[pos]], pos + 1, k + 1); }}void dfs2(ll res, int pos, int k) { if (res > sum)return; if (k > K)return; if (pos > n) { b[pli(res, k)] ++; return; } dfs2(res, pos + 1, k); dfs2(res + val[pos], pos + 1, k); if (val[pos] <= 20) { dfs2(res + fac[val[pos]], pos + 1, k + 1); }}int main(){// ios::sync_with_stdio(0); fac[1] = fac[0] = 1ll; for (int i = 2; i <= 20; i++)fac[i] = fac[i - 1] * i; cin >> n >> K >> sum; for (int i = 1; i <= n; i++)rdint(val[i]); dfs1(0, 1, 0); dfs2(0, n / 2 + 1, 0); map
                             
                              ::iterator it; for (it = a.begin(); it != a.end(); it++) { int j = (*it).first.second; for (int i = 0; i + j <= K; i++) { if (b.count(make_pair(sum - (*it).first.first, i))) { ans += 1ll * (*it).second*(b[pli(sum - (*it).first.first, i)]); } } } cout << ans * 1ll << endl; return 0;}

转载于:https://www.cnblogs.com/zxyqzy/p/10358054.html

你可能感兴趣的文章