BZOJ - 2844 线性基

题意:求给定的数在原数组中的异或组合中的排名(非去重)

因为线性基中\(b[j]=1\)表示该位肯定存在,所以给定的数如果含有该位,由严格递增和集合枚举可得,排名必然加上\(2^j\)(不是完全对角就需要额外维护),但这是去重后的结果

可证明的结论是每个数都重复出现了\(2^{n-|B|}\)次

#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 = 3e5+11;

const double EPS = 1e-7;

typedef long long ll;

typedef unsigned long long ull;

const ll MOD = 10086;

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,cnt;

vector<ll> vec;

void cal(int n){

    memset(b,0,sizeof b);cnt=0;

    rep(i,1,n){

        rrep(j,62,0){

            if(a[i]>>j&1){

                if(b[j]) a[i]^=b[j];

                else{

                    b[j]=a[i];cnt++;

                    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;

                }

            }

        }

    }

}

ll fpw(ll a,ll n,ll mod){

    ll res=1;

    while(n){

        if(n&1) res=(res*a)%mod;

        n>>=1;a=(a*a)%mod;

    }

    return res;

}

int main(){

    while(cin>>n){

        rep(i,1,n) a[i]=read();

        cal(n);

        vec.clear();

        rep(i,0,62) if(b[i]) vec.push_back(i);

        ll x=read(), rk=0;

        for(int i=0;i<vec.size();i++) if(x>>vec[i]&1) rk|=(1<<i);

        ll res=rk%MOD*fpw(2,n-cnt,MOD)%MOD+1;

        println(res%MOD);

    }

    return 0;

}