XOR

Consider an array A with n elements. Each of its element is A[i] (1 ≤ i ≤ n). Then gives two integers
Q, K, and Q queries follow. Each query, give you L, R, you can get Z by the following rules.
To get Z, at first you need to choose some elements from A[L] to A[R], we call them A[i1], A[i2],
. . . , A[it], Then you can get number Z = K or (A[i1], A[i2], . . . , A[it]).
Please calculate the maximum Z for each query .

Input
Several test cases.
First line an integer T (1 ≤ T ≤ 10). Indicates the number of test cases.
Then T test cases follows. Each test case begins with three integer N, Q, K (1 ≤ N ≤ 10000,1 ≤
Q ≤ 100000, 0 ≤ K ≤ 100000). The next line has N integers indicate A[1] to A[N] (0 ≤ A[i] ≤ 108).
Then Q lines, each line two integer L, R (1 ≤ L ≤ R ≤ N).

Output
For each query, print the answer in a single line.

Sample Input
1
5 3 0
1 2 3 4 5
1 3
2 4
3 5

Sample Output
3
7
7

題意：給出一個1個長度為n的數組A,然後給出q個詢問對於每個詢問，每次在下標為[l,r]的數中，選取一部分數，使得其異或值OR上k後最大，輸出這個最大值

分析：根據題意，能夠想到需要用到線性基。然後因為是要求最後OR上k後最大，根據OR運算的性質，可以先將數組中的數的二進制位上，將k為1的位置都變為0.因為這些位置在最後OR上k後都會變為1，所以暫時不討論它。比如現在數x為5(101)，k為4(100),那麽需要將x變為1.

轉換完成後，線性基的最大值OR上k就是答案，又因為題目是區間查詢，所以需要用到線段樹來維護線性基

代碼如下：

#include <cstdio>
#include 
 
<iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int MAXN=1e4+100;
typedef long long LL;
LL t,n,q,k,l,r,pd;
int vis[35];

long long read(){
    long long res = 0;
    int flag = 0;
    char ch;
    if ((ch = getchar()) == ‘-‘){
        flag = 1;
    }
    else
 
 if(ch >= ‘0‘ && ch <= ‘9‘){
        res = ch - ‘0‘;
    }
    while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘){
        res = res * 10 + (ch - ‘0‘);
    }

    return flag ? -res : res;
}

struct L_B{
    long long d[61];
    int cnt;
    L_B()
    {
        memset(d,0,sizeof(d));
        cnt=0;
    }
    void clear()
    {
        memset(d,0,sizeof(d));
        cnt=0;
    }
    void insert(long long val)
    {
        for (int i=31;i>=0;i--)
            if (val&(1LL<<i))
            {
                if (!d[i])
                {
                    d[i]=val;
                    break;
                }
                val^=d[i];
            }
    }
    long long query_max()
    {
        long long ret=0;
        for (int i=31;i>=0;i--)
        {
            if ((ret^d[i])>ret)
                 ret^=d[i];
        }
        return ret;
    }
};

L_B R;
L_B merge(const L_B &n1,const L_B &n2)
{
    L_B ret=n1;
    for (int i=31;i>=0;i--)
        if (n2.d[i])
            ret.insert(n2.d[i]);
    return ret;
}

struct node
{
    int l;
    int r;
    L_B A;
}tree[MAXN<<2];
void PushUp(int rt)
{
  tree[rt].A=merge(tree[rt<<1].A,tree[rt<<1|1].A);
}
void BuildTree(int l,int r,int rt)
{
    tree[rt].l=l;
    tree[rt].r=r;
    if(l==r)
    {
      LL x;
      x=read();
      x=x&pd;//轉換後，再插入到線性基中
      tree[rt].A.clear();
      tree[rt].A.insert(x);
      return;
    }
    int mid=(tree[rt].l+tree[rt].r)/2;
    BuildTree(l,mid,rt<<1);
    BuildTree(mid+1,r,rt<<1|1);
    PushUp(rt);
}
void Query(int l,int r,int rt)
{
    if(tree[rt].l==l&&tree[rt].r==r)
    {
      R=merge(R,tree[rt].A);
      return;
    }
    int mid=(tree[rt].l+tree[rt].r)/2;
    if(r<=mid)Query(l,r,rt<<1);
    else if(l>mid)Query(l,r,rt<<1|1);
    else
    {
        Query(l,mid,rt<<1);
        Query(mid+1,r,rt<<1|1);
    }
}

int main()
{
    t=read();
    while(t--)
    {
        pd=0;
        memset(vis,0,sizeof(vis));
       n=read(),q=read(),k=read();
        for (int i=31;i>=0;i--)
        {
          if(k&(1LL<<i))
          vis[i]=1;
          else
          pd+=(1LL<<i);// 用來進行轉換,pd&x即為轉換後的值
        }
        BuildTree(1,n,1);
        while(q--)
        {
           l=read(),r=read();
            R.clear();
            Query(l,r,1);
            LL ans=(R.query_max()|k);
            printf("%lld\n",ans);
        }
    }
    return 0;
}

2017 ICPC 西安站現場賽 A.XOR （線段樹+線性基）