Xor Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others) Total Submission(s): 1427    Accepted Submission(s): 595 Problem Description Zeus 和 Prometheus 做了一个游戏，Prometheus 给 Zeus 一个集合，集合中包含了N个正整数，随后 Prometheus 将向 Zeus 发起M次询问，每次询问中包含一个正整数 S ，之后 Zeus 需要在集合当中找出一个正整数 K ，使得 K 与 S 的异或结果最大。Prometheus 为了让 Zeus 看到人类的伟大，随即同意 Zeus 可以向人类求助。你能证明人类的智慧么？   Input 输入包含若干组测试数据，每组测试数据包含若干行。 输入的第一行是一个整数T（T < 10），表示共有T组数据。 每组数据的第一行输入两个正整数N，M（<1=N,M<=100000），接下来一行，包含N个正整数，代表 Zeus 的获得的集合，之后M行，每行一个正整数S，代表 Prometheus 询问的正整数。所有正整数均不超过2^32。   Output 对于每组数据，首先需要输出单独一行”Case #?:”，其中问号处应填入当前的数据组数，组数从1开始计算。 对于每个询问，输出一个正整数K，使得K与S异或值最大。   Sample Input 2 3 2 3 4 5 1 5 4 1 4 6 5 6 3   Sample Output Case #1: 4 3 Case #2: 4   Source 2014年百度之星程序设计大赛 - 资格赛  

对Trie树理解不到位，照着jhz033的套路写了

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 2000005; int a[N][3]; int sz; void Init() { memset(a,0,sizeof(a)); sz = 1; } void Insert(int x) { int num[35]; memset(num,0,sizeof(num)); int cnt = 0; while(x) { num[cnt++] = x % 2; x /= 2; } int u = 0,n = 34; for(int i = n;i >= 0;i--) { if(!a[u][num[i]]) a[u][num[i]] = sz++; u = a[u][num[i]]; } } int Query(int y) { int num[35]; memset(num,0,sizeof(num)); int cnt = 0; while(y) { num[cnt++] = y % 2; y /= 2; } for(int i = 0;i <= 34;i++) num[i]= num[i] ? 0 : 1; int u = 0,n = 34,v,w; int ans = 0; for(int i = n;i >= 0;i--) { if(num[i]) { v = 1; w = 0; } else { w = 1; v = 0; } if(a[u][v]) { u = a[u][v]; if(v) ans += 1 << i; } else { u = a[u][w]; if(!v) ans += 1 << i; } } return ans; } int main() { int t,n,m; cin >> t; for(int ca = 1;ca <= t;ca++) { Init(); cin >> n >> m; int x,y; for(int i = 1;i <= n;i++) { scanf("%d",&x); Insert(x); } printf("Case #%d:\n",ca); for(int i = 1;i <= m;i++) { scanf("%d",&y); printf("%d\n",Query(y)); } } return 0; } Codeforces Round #367 (Div. 2)

D. Vasiliy's Multiset time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output

Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

"+ x" — add integer x to multiset A. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query. "? x" — you are given integer x and need to compute the value , i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.

Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer x_i (1 ≤ x_i ≤ 10⁹). It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer x_i and some integer from the multiset A.

Example input 10 + 8 + 9 + 11 + 6 + 1 ? 3 - 8 ? 3 ? 8 ? 11 output 11 10 14 13 Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer  — maximum among integers , , ,  and .

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 4000005; int a[N][3],sum[N]; int sz; void Init() { memset(a,0,sizeof(a)); memset(sum,0,sizeof(sum)); sz = 1; } void Insert(int x) { int num[35]; memset(num,0,sizeof(num)); int cnt = 0; while(x) { num[cnt++] = x % 2; x /= 2; } int u = 0,n = 34; for(int i = n;i >= 0;i--) { if(!a[u][num[i]]) a[u][num[i]] = sz++; u = a[u][num[i]]; sum[u]++; } } void Delete(int x) { int num[35]; memset(num,0,sizeof(num)); int cnt = 0; while(x) { num[cnt++] = x % 2; x /= 2; } int u = 0,n = 34; for(int i = n;i >= 0;i--) { int v = a[u][num[i]]; sum[v]--; if(!sum[v]) a[u][num[i]] = 0; u = v; } } int Query(int y) { int num[35]; memset(num,0,sizeof(num)); int cnt = 0; while(y) { num[cnt++] = y % 2; y /= 2; } for(int i = 0;i <= 34;i++) num[i]= num[i] ? 0 : 1; int u = 0,n = 34,v,w; int ans = 0; for(int i = n;i >= 0;i--) { if(num[i]) { v = 1; w = 0; } else { w = 1; v = 0; } if(a[u][v]) { u = a[u][v]; ans += 1 << i; } else { u = a[u][w]; //if(!v) //ans += 1 << i; } } return ans; } int main() { int t,x; cin >> t; char s[20]; Init(); Insert(0); for(int ca = 1;ca <= t;ca++) { scanf("%s%d",s,&x); if(s[0] == '+') Insert(x); else if(s[0] == '-') Delete(x); else printf("%d\n",Query(x)); } return 0; } 感觉这样的代码有点丑陋，但是原理应该一样，弄熟了回来改