LibreOJ #114. k 大异或和

二次联通门 : LibreOJ #114. k 大异或和

/*
    LibreOJ #114. k 大异或和

    WA了很多遍
    为什么呢。。。

    一开始读入原数的时候写的是for(;N--;)
    而重新构造线性基的时候要用到N。。。所以GG

    对于找第k大异或和
    只需把原来的线性基重新构造
    构造规则为
        若i<j， aj的第j位是1，就把aj异或上ai

    查询的时候将k二进制拆分，对于1的位，就异或上对应的线性基。
    最终得出的答案就是k小值。
*/
#include <cstring>
#include <cstdio>

inline void read (long long &now)
{
    register char word = getchar ();
    bool temp = false;
    for (now = 0; word < ‘0‘ || word > ‘9‘; word = getchar ())
        if (word == ‘-‘)
            temp = true;
    for (; word >= ‘0‘ && word <= ‘9‘; now = now * 10 + word - ‘0‘, word = getchar ());
    if (temp)
        now = -now;
}

long long N, M;

class Linear_Base_Type
{
    static const int _L = 52;

    private :

        long long number[_L + 2];
        long long data[_L + 2];

        int Count;

    public :

        Linear_Base_Type ()
        {
            memset (data, 0, sizeof data);
            memset (number, 0, sizeof number);
            Count = 0;
        }

        inline void Insert (register long long key)
        {
            for (register int i = _L; i >= 0; i --)
                if (key & (1LL << i))
                {
                    if (number[i] == 0)
                    {
                        number[i] = key;
                        break;
                    }
                    key ^= number[i];
                }
            return ;
        }

        void Re_Build ()
        {
            for (register int i = 1, j; i <= _L; i ++)
                for (j = 0; j < i; j ++)
                    if ((1LL << j) & number[i])
                        number[i] ^= number[j];

            for (register int i = 0; i <= _L; i ++)
                if (number[i])
                    data[Count ++] = number[i];
        }

        inline long long Query_kth (register long long k)
        {
            long long res = 0;

            if (Count != N)
                -- k;
            if (k >= (1LL << Count))
                return -1;

            for (register int i = 0; i <= _L; i ++)
                if (k & (1LL << i))
                    res ^= data[i];

            return res;
        }
};

Linear_Base_Type Make;

int main (int argc, char *argv[])
{
    long long x;
    read (N);

    for (int i = 1; i <= N; i ++)
    {
        read (x);

        Make.Insert (x);
    }
    read (M);
    for (Make.Re_Build (); M --; )
    {
        read (x);

        printf ("%lld\n", Make.Query_kth (x));
    }
    return 0;
}