Codeforces Round #430 (Div. 2) D.Vitya and Strange Lesson 异或 01字典树补集最小
2017-09-02 21:48
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D. Vitya and Strange Lesson
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Today at the lesson Vitya learned a very interesting function — mex. Mex of a sequence of numbers is the minimum non-negative number
that is not present in the sequence as element. For example, mex([4, 33, 0, 1, 1, 5]) = 2 and mex([1, 2, 3]) = 0.
Vitya quickly understood all tasks of the teacher, but can you do the same?
You are given an array consisting of n non-negative integers, and m queries.
Each query is characterized by one number x and consists of the following consecutive steps:
Perform the bitwise addition operation modulo 2 (xor) of each array element
with the number x.
Find mex of the resulting array.
Note that after each query the array changes.
Input
First line contains two integer numbers n and m (1 ≤ n, m ≤ 3·105) —
number of elements in array and number of queries.
Next line contains n integer numbers ai (0 ≤ ai ≤ 3·105) —
elements of then array.
Each of next m lines contains query — one integer number x (0 ≤ x ≤ 3·105).
Output
For each query print the answer on a separate line.
Examples
input
output
input
output
input
output
题意:求区间没有的最小的那个数,可以异或x。
思路:首先明确对[1,2^n]区间内的所有数亦或x,仍然是[1,2^n]的所有数。
设给定集合为A,全集C,A的补集为B。
易知每次亦或A无需真的改动A数组,A^x1^x2 = A^(x1^x2)。
每次query对A中每个数亦或x, 然后查询min(C-A^x)。
又A^x+B^x = A + B = C,得所求即为min(B^x)。用字典树来求解亦或最值问题。
D. Vitya and Strange Lesson
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Today at the lesson Vitya learned a very interesting function — mex. Mex of a sequence of numbers is the minimum non-negative number
that is not present in the sequence as element. For example, mex([4, 33, 0, 1, 1, 5]) = 2 and mex([1, 2, 3]) = 0.
Vitya quickly understood all tasks of the teacher, but can you do the same?
You are given an array consisting of n non-negative integers, and m queries.
Each query is characterized by one number x and consists of the following consecutive steps:
Perform the bitwise addition operation modulo 2 (xor) of each array element
with the number x.
Find mex of the resulting array.
Note that after each query the array changes.
Input
First line contains two integer numbers n and m (1 ≤ n, m ≤ 3·105) —
number of elements in array and number of queries.
Next line contains n integer numbers ai (0 ≤ ai ≤ 3·105) —
elements of then array.
Each of next m lines contains query — one integer number x (0 ≤ x ≤ 3·105).
Output
For each query print the answer on a separate line.
Examples
input
2 2 1 3 1 3
output
1 0
input
4 3 0 1 5 6 1 2 4
output
2 0 0
input
5 4 0 1 5 6 7 1 1 4 5
output
2 2 0 2
题意:求区间没有的最小的那个数,可以异或x。
思路:首先明确对[1,2^n]区间内的所有数亦或x,仍然是[1,2^n]的所有数。
设给定集合为A,全集C,A的补集为B。
易知每次亦或A无需真的改动A数组,A^x1^x2 = A^(x1^x2)。
每次query对A中每个数亦或x, 然后查询min(C-A^x)。
又A^x+B^x = A + B = C,得所求即为min(B^x)。用字典树来求解亦或最值问题。
//china no.1 #pragma comment(linker, "/STACK:1024000000,1024000000") #include <vector> #include <iostream> #include <string> #include <map> #include <stack> #include <cstring> #include <queue> #include <list> #include <stdio.h> #include <set> #include <algorithm> #include <cstdlib> #include <cmath> #include <iomanip> #include <cctype> #include <sstream> #include <functional> #include <stdlib.h> #include <time.h> #include <bitset> using namespace std; #define pi acos(-1) #define PI acos(-1) #define endl '\n' #define srand() srand(time(0)); #define me(x,y) memset(x,y,sizeof(x)); #define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++) #define close() ios::sync_with_stdio(0); cin.tie(0); #define FOR(x,n,i) for(int i=x;i<=n;i++) #define FOr(x,n,i) for(int i=x;i<n;i++) #define fOR(n,x,i) for(int i=n;i>=x;i--) #define fOr(n,x,i) for(int i=n;i>x;i--) #define W while #define sgn(x) ((x) < 0 ? -1 : (x) > 0) #define bug printf("***********\n"); #define db double #define ll long long typedef long long LL; const int INF=0x3f3f3f3f; const LL LINF=0x3f3f3f3f3f3f3f3fLL; const int dx[]={-1,0,1,0,1,-1,-1,1}; const int dy[]={0,1,0,-1,-1,1,-1,1}; const int maxn=3e5+10; const int maxx=1e6+100; const double EPS=1e-8; const double eps=1e-8; const int mod=1e9+7; template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);} template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);} template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));} template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));} template <class T> inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;} while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;} inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false; while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;} else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}} if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}} if(IsN) num=-num;return true;} void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');} void print(LL a){ Out(a),puts("");} //freopen( "in.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); //cerr << "run time is " << clock() << endl; int next[maxn*20][2]; int val[maxn*20]; int st; void init() { me(next[0],0); me(val,0); st=1; } void insert(LL x) { int u=0; for(int i=20;i>=0;i--) { int c=((x>>i)&1); if(!next[u][c]) { me(next[st],0); next[u][c]=st++; } u=next[u][c]; ++val[u]; } } LL query(LL x) { int t=0; LL ans=0; for(int i=20;i>=0;i--) { int k=((x>>i)&1); if(!next[t][k]) return ans; if(val[next[t][k]]==(1<<i)) { ans|=(1<<i); t=next[t][1-k]; } else t=next[t][k]; } return ans; } int n,m,vis[maxn]; int main() { scan_d(n); scan_d(m); init(); int x; FOR(1,n,i) { scan_d(x); if(vis[x]) continue; else { vis[x]=1; insert(x); } } int y=0; FOR(1,m,i) { scan_d(x); y^=x; print(query(y)); } }
#include <iostream> #include <string> #include <vector> #include <stack> #include <queue> #include <deque> #include <set> #include <map> #include <algorithm> #include <functional> #include <utility> #include <cstring> #include <cstdio> #include <cstdlib> #include <ctime> #include <cmath> #include <cctype> #define CLEAR(a, b) memset(a, b, sizeof(a)) #define IN() freopen("in.txt", "r", stdin) #define OUT() freopen("out.txt", "w", stdout) #define LL long long #define maxn 524288 #define maxm 6000005 #define mod 10007 #define INF 1000000007 #define EPS 1e-7 #define PI 3.1415926535898 #define N 4294967296 using namespace std; //-------------------------CHC------------------------------// struct Node { Node *next[2]; Node() { next[0] = next[1] = NULL; } }node[maxm]; int total; void insert(int num, Node *root) { Node *cur = root; for (int i = 19; i >= 0; --i) { int id = num >> i & 1; if (!cur->next[id]) cur->next[id] = &node[total++]; cur = cur->next[id]; } } int query(int num, Node *root) { int ret = 0; Node *cur = root; for (int i = 19; i >= 0; --i) { int id = num >> i & 1; if (!cur->next[id]) id = !id, ret |= (1 << i); cur = cur->next[id]; } return ret; } bool vis[maxn+10]; int main() { int n, q; scanf("%d%d", &n, &q); int x; for (int i = 0; i < n; ++i) scanf("%d", &x), vis[x] = 1; Node *root = &node[total++]; for (int i = 0; i <= maxn; ++i) if (!vis[i]) insert(i, root); int y = 0; while (q--) { scanf("%d", &x); y ^= x; printf("%d\n", query(y, root)); } return 0; }
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