[bzoj4800][Ceoi2015]Ice Hockey World Championship

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n<=40 m<=10^18

考虑双向宽搜，然后得到两个大小为2^20的数组，排序之后两个指针推一推计算答案即可。

排序最好用基数排序

#include<iostream>
#include<cstdio>
#include<cstring>
#define MN 40
#define MM 1050000
#define N 32767
#define ll unsigned long long
using namespace std;
inline ll read()
{
    ll x = 0 , f = 1; char ch = getchar();
    while(ch < ‘0‘ || ch > ‘9‘){ if(ch == ‘-‘) f = -1;  ch = getchar();}
    while(ch >= ‘0‘ && ch <= ‘9‘){x = x * 10 + ch - ‘0‘;ch = getchar();}
    return x * f;
}

int n,lim,tot1=0,tot2=0,v[4][N+5];
ll a[MN+5],s[MM+5],s2[MM+5],m,sa[MM+5],ans=0;

void Dfs2(int x,ll y)
{
    if(x==lim){s2[++tot2]=y;return;}
    if(y+a[x]<=m) Dfs2(x-1,y+a[x]);
    Dfs2(x-1,y);
}

void Dfs1(int x,ll y)
{
    if(x>lim){s[++tot1]=y;return;}
    if(y+a[x]<=m)Dfs1(x+1,y+a[x]);
    Dfs1(x+1,y);
}

void Sort(ll*s,int n)
{
    memset(v,0,sizeof(v));
//    for(int i=1;i<=n;++i) cout<<s[i]<<" ";puts("");
    for(int i=1;i<=n;++i)
        for(ll j=s[i],i=0;i<4;j>>=15,++i) ++v[i][j&N];
    for(int r=0;r<4;++r) for(int i=1;i<=N;++i) v[r][i]+=v[r][i-1];
    for(ll r=0,j=N;r<4;++r,j<<=15)
    {
        for(register int i=n;i;--i) sa[v[r][(s[i]&j)>>(r*15)]--]=s[i];
        for(register int i=1;i<=n;++i) s[i]=sa[i];
    }
}

int main()
{
    n=read();lim=n>>1;m=read();
    for(int i=1;i<=n;++i) a[i]=read();
    Dfs1(1,0);Dfs2(n,0);
    Sort(s,tot1);Sort(s2,tot2);
    for(int i=tot1,j=1;i;--i)
    {
        while(j<tot2&&s2[j+1]+s[i]<=m) ++j;
        if(s2[j]+s[i]<=m) ans+=j;
    }
    printf("%lld\n",ans);
    return 0;
}