BZOJ 4800([Ceoi2015]Ice Hockey World Championship-meet in the middle)

Description

有n个物品，m块钱，给定每个物品的价格，求买物品的方案数。
Input

第一行两个数n，m代表物品数量及钱数
第二行n个数，代表每个物品的价格
n<=40，m<=10^18
Output

一行一个数表示购买的方案数
（想怎么买就怎么买，当然不买也算一种）
Sample Input

5 1000

100 1500 500 500 1000
Sample Output

8
HINT

只有40个物品的0.1背包
我们双向广搜，然后排序合并。
复杂度O(2nlog(2n))=O(2n∗n)

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;}#define MAXN (6448576)int n,cnt=0; ll a[MAXN],v[MAXN];ll m;void dfs(int x,ll s,int n) {    if (s>m) return;     if (x==n+1) {        v[++cnt]=s;        return;    }    dfs(x+1,s+a[x],n);    dfs(x+1,s,n);}ll Ans=0;void dfs2(int x,ll s,int n) {    if (s>m) return;     if (x==n+1) {        int l=1,r=cnt,ans=0;        while(l<=r) {            int mi=(l+r)/2;            if (s+v[mi]<=m) ans=mi,l=mi+1;else r=mi-1;        }        Ans+=ans;        return;    }    dfs2(x+1,s+a[x],n);    dfs2(x+1,s,n);}int main(){//  freopen("bzoj4800.in","r",stdin);//  freopen(".out","w",stdout);    n=read();    cin>>m;    For(i,n) cin>>a[i];    int mid=n/2;    dfs(1,0,mid);    sort(v+1,v+1+cnt);    dfs2(mid+1,0,n);    cout<<Ans<<endl;    return 0;}