Find MaxXorSum
Time Limit:2000MS Memory Limit:65535KB 64bit IO Format:
Description
Given n non-negative integers, you need to find two integers a and b that a xor b is maximum. xor is exclusive-or.
Input
Input starts with an integer T(T <= 10) denoting the number of tests.
For each test case, the first line contains an integer n(n <= 100000), the next line contains a1, a2, a3, ......, an(0 <= ai <= 1000000000);
Output
For each test case, print you answer.
Sample Input
2
4
1 2 3 4
3
7 12 5
Sample Output
7
11
这个题目其实思路很简单，就我知道的写法有Trie树指针和静态数组两种写法，一开始写的指针用了pow函数TLE了2次，然后换位运算又TLE两次醉了。。
后面换成静态数组数组开大了RE了一次。。。哭。。。
这是AC代码：
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1 #include"iostream"
2 #include"algorithm"
3 #include"cstdio"
4 #include"cmath"
5 #include"cstring"
6 #define MX 1400000
7 #define INF 0x3f3f3f3f
8 #define lson l,m,rt<<1
9 #define rson m+1,r,rt<<1|1
10 using namespace std;
11 int tree[MX][2],xx;
12
13 void BuildTrie(long long a) {
14 int i=31;
15 int p=0;
16 while(0<=i) {
17 bool num=a&(1<<i);
18 if(!tree[p][num]) { //如果节点为空
19 tree[p][num]=++xx;//标记并创建新的子节点
20 }
21 p=tree[p][num];
22 // cout<<"YES"<<i<<" B is:"<<num<<"\n"; //建立的Trie树的样子
23 i--;
24 }
25 }
26
27 long long Query(long long a) {
28 int i=31,p=0;
29 long long ans=0;
30 while(0<=i) {
31 bool num=!(a&(1<<i)); //取反查找
32 if(tree[p][num]) p=tree[p][num],ans+=1<<i;//如果和原来的那一为相反的存在的话，返回值就加上，并且在这个支路走
33 else p = tree[p][!num]; //按照相同的顺序找
34 // cout<<"YES"<<i<<" B is:"<<num<<" ans is:"<<ans<<endl; //查询时候的Trie树的样子
35 i--;
36 }
37 return ans;
38 }
39
40 int main() {
41 int T,n;
42 long long a[100050],ans;
43 scanf("%d",&T);
44 while(T--) {
45 ans=0;
46 memset(tree,0,sizeof(tree));
47 xx=0;
48 scanf("%d",&n);
49 for(int i=1; i<=n; i++) {
50 scanf("%I64d",&a[i]);
51 BuildTrie(a[i]);
52 }
53 //cout<<"YES1\n";
54 for(int i=1; i<=n; i++) {
55 ans=max(ans,Query(a[i]));
56 }
57 // cout<<"YES2\n";
58 printf("%I64d\n",ans);
59 }
60 return 0;
61 }
View Code
这是指针的写法，感觉复杂度和上面的没啥区别，为啥就ＴＬＥ了呢？ 希望有人知道给我指点下。
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1 #include"iostream"
2 #include"algorithm"
3 #include"cstdio"
4 #include"cmath"
5 #include"cstring"
6 #define MX 110000
7 #define INF 0x3f3f3f3f
8 #define lson l,m,rt<<1
9 #define rson m+1,r,rt<<1|1
10 using namespace std;
11
12 struct Trie {
13 Trie *next[2];
14 } root;
15
16 void BuildTrie(int a) {
17 Trie *p=&root,*q;
18 int i=31;
19 while(i>=0) {
20 bool num=a&(1<<i);
21 if(p->next[num]==NULL) {
22 q=(Trie *)malloc(sizeof(root));//申请一块新内存; //动态分配内存
23 q->next[1]=NULL;
24 q->next[0]=NULL; //清空申请内存的所有子节点
25 p->next[num]=q; //往子节点下去继续记录字典树
26 p=p->next[num];
27 } else {
28 p=p->next[num];
29 }
30 // cout<<"YES"<<i<<" B is:"<<num<<"\n"; //建立的Trie树的样子
31 i--;
32 }
33 }
34
35 long long Query(int a) {
36 Trie *p=&root;
37 long long ans=0;
38 int i=31;
39 while(0<=i) {
40 bool num=a&(1<<i);
41 if(p->next[!num]!=NULL) p=p->next[!num],ans+=1<<(i);//如果和原来的那一为相反的存在的话，返回值就加上，并且在这个支路走
42 else if(p->next[num]!=NULL) p=p->next[num]; //按照相同的顺序找
43 // cout<<"YES"<<i<<" B is:"<<num<<"\n"<<"ans is:"<<ans<<endl; //查询时候的Trie树的样子
44 i--;
45 }
46 return ans;
47 }
48
49 int main() {
50 int T,n,a[MX];
51 long long ans;
52 scanf("%d",&T);
53 while(T--) {
54 ans=0;
55 root.next[0]=NULL;
56 root.next[1]=NULL;
57 scanf("%d",&n);
58 for(int i=0; i<n; i++) {
59 scanf("%d",&a[i]);
60 BuildTrie(a[i]);
61 }
62 //cout<<"YES1\n";
63 for(int i=0; i<n; i++) {
64 ans=max(ans,Query(a[i]));
65 }
66 // cout<<"YES2\n";
67 printf("%I64d\n",ans);
68 }
69 return 0;
70 }
View Code