HDU-3949 Linear Basis Applications

The meaning of the question: find the xor sum of the smallest \(k\)

Key points:
1. The linear basis can represent any XOR sum of the original array, but does not include 0, which requires special judgment (flag)
2. The XOR combination in the linear basis is only \(2^{|B|}-1\) If it can be XORed to 0, then the number of combinations is \(2^{|B|}\) 3.
The linear basis will inevitably increase after removing the 0 in the upper triangular matrix, that is, \(2^{|B |}-1\) is strictly increasing, so the binary value according to \(k\) is
inevitable

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<bitset>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar('\n')
#define blank putchar(' ')
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int MAXN = 1e6+11;
const double EPS = 1e-7;
typedef long long ll;
typedef unsigned long long ull;
const ll MOD = 1e9+7; 
unsigned int SEED = 17;
const ll INF = 1ll<<60;
ll read(){
    ll x=0,f=1;register 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;
}
ll a[MAXN],b[66],n;
bool flag=1;
ll cal(){
    memset(b,0,sizeof b);
    flag=1;
    rep(i,1,n){
        rrep(j,62,0){
            if(a[i]>>j&1){
                if(b[j]) {a[i]^=b[j];flag&=bool(a[i]>0);}
                else{
                    b[j]=a[i];
                    rrep(k,j-1,0) if(b[k]&&(b[j]>>k&1))b[j]^=b[k];
                    rep(k,j+1,62) if(b[k]>>j&1) b[k]^=b[j];
                    break;
                }
            }
        }
    }
    sort(b,b+62+1);
    return unique(b,b+63)-b;
}

ll query(int k){
    int cur=b[0]==0?1:0;
    ll ans=0;
    while(k){
        ans^=(b[cur++]*(k&1));
        k>>=1;
    }
    return ans;
}
int main(){
    int T=read(),kase=0;
    while(T--){
        n=read();
        rep(i,1,n) a[i]=read();
        int tot=cal();flag^=1;
        int q=read();
        printf("Case #%d:\n",++kase);
        rep(i,1,q){
            ll k=read();if(flag)k--;
            if(k==0) println(0);
            else if(k>=(1ll<<(tot-bool(b[0]==0)))) println(-1);
            else println(query(k));
        }
    }
    return 0;
}