算法笔记--线性基求交模板

https://blog.csdn.net/demon_rieman/article/details/88830846
模板1：

struct LinearBase {
    LL v[32];
    inline void clr() {memset(v, 0, sizeof(v));}
    inline void ins(LL a) {
        for (int i = 31; i >= 0; --i) {
            if(a&(1LL<<i)) {
                if(!v[i]) {
                    v[i] = a;
                    break;
                }
                a ^= v[i];
            }
        }
    }
    inline bool can_ex(LL a) {
        for (int i = 31; i >= 0; --i) if((a^v[i]) < a)  a ^= v[i];
        return a == 0;
    }
    //交
    inline friend LinearBase operator * (const LinearBase &a, const LinearBase &b) {
        LinearBase all, c, d;
        all.clr(), c.clr(), d.clr();
        for (int i = 31; i >= 0; --i) {
            all.v[i] = a.v[i];
            d.v[i] = 1LL<<i;
        }
        for (int i = 31; i >= 0; --i) {
            if(b.v[i]) {
                LL v = b.v[i], k = 0;
                bool can = true;
                for (int j = 31; j >= 0; --j) {
                    if(v&(1LL<<j)) {
                        if(all.v[j]) {
                            v ^= all.v[j];
                            k ^= d.v[j];
                        }
                        else {
                            can = false;
                            all.v[j] = v;
                            d.v[j] = k;
                            break;
                        }
                    }
                }
                if(can) {
                    LL v = 0;
                    for (int j = 31; j >= 0; --j) {
                        if(k&(1LL<<j)) {
                            v ^= a.v[j];
                        }
                    }
                    c.ins(v);
                }
            }
        }
        return c;
    }
};

模板2：

struct LinearBase {
    LL v[32];
    inline void clr() {memset(v, 0, sizeof(v));}
    inline void ins(LL a) {
        for (int i = 31; i >= 0; --i) {
            if(a&(1LL<<i)) {
                if(!v[i]) {
                    v[i] = a;
                    break;
                }
                a ^= v[i];
            }
        }
    }
    inline bool can_ex(LL a) {
        for (int i = 31; i >= 0; --i) if((a^v[i]) < a)  a ^= v[i];
        return a == 0;
    }
    //交
    inline friend LinearBase operator * (const LinearBase &a, const LinearBase &b) {
        LinearBase all, c, d;
        all.clr(), c.clr(), d.clr();
        for (int i = 31; i >= 0; --i) all.v[i] = a.v[i];
        for (int i = 31; i >= 0; --i) {
            if(b.v[i]) {
                LL v = b.v[i], k = 1LL<<i;
                bool can = true;
                for (int j = 31; j >= 0; --j) {
                    if(v&(1LL<<j)) {
                        if(all.v[j]) {
                            v ^= all.v[j];
                            k ^= d.v[j];
                        }
                        else {
                            can = false;
                            all.v[j] = v;
                            d.v[j] = k;
                            break;
                        }
                    }
                }
                if(can) {
                    LL v = 0;
                    for (int j = 31; j >= 0; --j) {
                        if(k&(1LL<<j)) {
                            v ^= b.v[j];
                        }
                    }
                    c.ins(v);
                }
            }
        }
        return c;
    }
}；