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 hasncubes lying in a line and numbered from1tonfrom left to right, with natural numbers written on them. She also haskstickers 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 reads5, then after the sticking it reads5!, which equals120.
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 mostkexclamation marks so that the sum of the numbers written on the chosen cubes after the sticking becomes equal toS. 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 integersn,kandS(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 containsnpositive integersai(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 numberS.
Sample test(s)
input
2 2 304 3
output
1
input
2 2 74 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.
题意：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。
思路：采用折半搜索，，自己太渣。。请看这位大神详解~
代码：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 1005#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b) for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i–)#define FRL(i,a,b) for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i–)#define mem(t, v) memset ((t) , v, sizeof(t))#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 pfprintf#define DBGpf("Hi\n")typedef __int64 ll;using namespace std;typedef pair<ll,ll>P;map<P,ll>a,b;ll n,k,s;ll p[30];ll val[30];void init(){int i;p[0]=p[1]=1;FRE(i,2,20)p[i]=p[i-1]*i;}void dfs(ll pos,ll num,ll kk){if (num>s||kk>k) return ;if (pos>n/2){a[P(num,kk)]++;return ;}dfs(pos+1,num+val[pos],kk);dfs(pos+1,num,kk);if (val[pos]<=20)dfs(pos+1,num+p[val[pos]],kk+1);return ;}void DFS(ll pos,ll num,ll kk){if (num>s||kk>k) return ;if (pos>n){b[P(num,kk)]++;return ;}DFS(pos+1,num+val[pos],kk);DFS(pos+1,num,kk);if (val[pos]<=20)DFS(pos+1,num+p[val[pos]],kk+1);return ;}int main(){ll i,j;init();while (~scanf("%I64d%I64d%I64d",&n,&k,&s)){a.clear();b.clear();FRE(i,1,n)scanf("%I64d",&val[i]);dfs(1,0,0);DFS(n/2+1,0,0);ll ans=0;for (map<P,ll>::iterator it=a.begin();it!=a.end();it++){int aa=(*it).first.second;for (int bb=0;aa+bb<=k;bb++){if (b.count(make_pair(s-(it->first.first),bb)))ans+=(it->second*b[make_pair(s-(it->first.first),bb)]);}}pf("%I64d\n",ans);}return 0;}
不会因为忧伤而风情万种。
