Chip Factory

Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 620 Accepted Submission(s): 318

Problem Description

John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces n chips today, the i-th chip produced this day has a serial number si.

At the end of the day, he packages all the chips produced this day, and send it to wholesalers. More specially, he writes a checksum number on the package, this checksum is defined as below:
maxi,j,k(si+sj)⊕sk

which i,j,k are three different integers between 1 and n. And ⊕ is symbol of bitwise XOR.

Can you help John calculate the checksum number of today?

Input
The first line of input contains an integer T indicating the total number of test cases.

The first line of each test case is an integer n, indicating the number of chips produced today. The next line has n integers s1,s2,..,sn, separated with single space, indicating serial number of each chip.

1≤T≤1000
3≤n≤1000
0≤si≤10⁹
There are at most 10 testcases with n>100

Output

For each test case, please output an integer indicating the checksum number in a line.

Sample Input

1 2 3

100 200 300

Sample Output

400

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

解题：很明显这种题目用字典树，今年的多校有一道类似的但是难度更大的题目

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn = ;

 struct Trie {

     int ch[maxn][],cnt[maxn],tot;

     int newnode() {

         cnt[tot] = ch[tot][] = ch[tot][] = ;

         return tot++;

     }

     void init() {

         tot = ;

         newnode();

     }

     void update(int v,int d,int rt = ) {

         for(int i = ; i >= ; --i) {

             int c = (v>>i)&;

             if(!ch[rt][c]) ch[rt][c] = newnode();

             rt = ch[rt][c];

             cnt[rt] += d;

         }

     }

     int match(int v,int rt = ,int ret = ) {

         for(int i = ; i >= ; --i) {

             int c = (v>>i)&;

             if(ch[rt][c^] && cnt[ch[rt][c^]]) {

                 ret |= <<i;

                 rt = ch[rt][c^];

             } else rt = ch[rt][c];

         }

         return ret;

     }

 }T;

 int a[maxn];

 int main() {

     int kase,n,ret;

     scanf("%d",&kase);

     while(kase--){

         scanf("%d",&n);

         T.init();

         for(int i = ; i < n; ++i){

             scanf("%d",a + i);

             T.update(a[i],);

         }

         for(int i = ret = ; i < n; ++i){

             T.update(a[i],-);

             for(int j = i + ; j < n; ++j){

                 T.update(a[j],-);

                 ret = max(ret,T.match(a[i] + a[j]));

                 T.update(a[j],);

             }

             T.update(a[i],);

         }

         printf("%d\n",ret);

     }

     return ;

 }

巴特西

HDU 5536 Chip Factory

Chip Factory

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