牛客练习赛77 - 小G的LY数对（hash表）

传送门：牛客练习赛77 - 小G的LY数对

思路分析

考虑meeting in the middle，枚举其中一位，查另一位的数量，减去自身重复的部分

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (int)(x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;

const int INF = 0x3f3f3f3f;
const int N = 3e5+10;

const int maxsz = 2e7+3;//@maxsz素数表@
//1e7+19,2e7+3,3e7+23,4e5+9 @maxsz最好为素数@
//1e6+3,2e6+3,3e6+7,4e6+9,1e5+3,2e5+3,3e5+7
//@要保证取值的操作值集合小于maxsz,@
//@count操作不增加新节点@
template<typename key,typename val>
class hash_map {
    public:
        struct node {
            key u;
            val v;
            int next;
        };
        vector<node> e;
        int head[maxsz],nume,numk,id[maxsz];
        bool count(key u) {
            int hs=(u%maxsz + maxsz) % maxsz;
            for(int i=head[hs]; i; i=e[i].next)
                if(e[i].u==u) return 1;
            return 0;
        }
        val& operator[](key u) {
            int hs=(u%maxsz + maxsz) % maxsz;
            for(int i=head[hs]; i; i=e[i].next)
                if(e[i].u==u) return e[i].v;
            if(!head[hs])id[++numk]=hs;
            if(++nume>=e.size())e.resize(nume<<1);
            return e[nume]=(node) {u,0,head[hs]},head[hs]=nume,e[nume].v;
        }
};

int a[N],b[N];

hash_map<int,int>cnt,cntxor;

signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n,m;
    read(n,m);
    for(int i=1; i<=n; i++) {
        read(a[i]);
        cnt[a[i]]++;
        for(int j=0; j<30; j++) {
            int x = a[i]^(1<<j);
            cntxor[x]++;
        }
    }
    ll ans = 0;
    for(int i=1; i<=m; i++) {
        int w;
        read(w);
        for(int j=0; j<30; j++) {
            int x = w^(1<<j);
            if(cntxor.count(x)) ans+=cntxor[x];
            if(cnt.count(w)) ans-=cnt[w];
        }
    }

    printf("%lld\n",ans/2);





    return 0;
}