标签：node har using include sample 区间 位置 AC 输出
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
题意：给出一个1个长度为n的数组A,然后给出q个询问对于每个询问，每次在下标为[l,r]的数中，选取一部分数，使得其异或值OR上k后最大，输出这个最大值
分析：根据题意，能够想到需要用到线性基。然后因为是要求最后OR上k后最大，根据OR运算的性质，可以先将数组中的数的二进制位上，将k为1的位置都变为0.因为这些位置在最后OR上k后都会变为1，所以暂时不讨论它。比如现在数x为5(101)，k为4(100),那么需要将x变为1.
转换完成后，线性基的最大值OR上k就是答案，又因为题目是区间查询，所以需要用到线段树来维护线性基
代码如下：
#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;//转换后，再插入到线性基中 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);// 用来进行转换,pd&x即为转换后的值 } 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 西安站现场赛 A.XOR （线段树+线性基）
标签：node har using include sample 区间 位置 AC 输出
原文地址：https://www.cnblogs.com/a249189046/p/8953588.html
