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 = ; word < '' || word > ''; word = getchar ())
        if (word == '-')
            temp = true;
    for (; word >= '' && word <= ''; now = now *  + word - '', word = getchar ());
    if (temp)
        now = -now;
}
 
long long N, M;
 
class Linear_Base_Type
{
    static const int _L = ;
 
    private :
 
        long long number[_L + ];
        long long data[_L + ];
 
        int Count;
 
    public :
 
        Linear_Base_Type ()
        {
            memset (data, , sizeof data);
            memset (number, , sizeof number);
            Count = ;
        }
 
        inline void Insert (register long long key)
        {
            for (register int i = _L; i >= ; i --)
                if (key & (1LL << i))
                {
                    if (number[i] == )
                    {
                        number[i] = key;
                        break;
                    }
                    key ^= number[i];
                }
            return ;
        }
 
        void Re_Build ()
        {
            for (register int i = , j; i <= _L; i ++)
                for (j = ; j < i; j ++)
                    if ((1LL << j) & number[i])
                        number[i] ^= number[j];
 
            for (register int i = ; i <= _L; i ++)
                if (number[i])
                    data[Count ++] = number[i];
        }
 
        inline long long Query_kth (register long long k)
        {
            long long res = ;
 
            if (Count != N)
                -- k;
            if (k >= (1LL << Count))
                return -;
 
            for (register int i = ; 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 = ; 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 ;
}