hdu 5536 Chip Factory (暴力)

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3227    Accepted Submission(s): 1393

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

231 2 33100 200 300

 

Sample Output

6400

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

这题正解是字典树。当时看到这题首先就是想到暴力了= =

思路：O(n^3)无疑是会TLE的，不过可以预处理一下降低到O(n*m)，先将（si+sj）这个值算出来存起来，然后再来算xor的值。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define MAX_N 1000000using namespace std;struct data{    int i,j,sum;}q[MAX_N];int a[1005];int main(){    int t,n,cnt;    scanf("%d",&t);    while(t--)    {        cnt=0;        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        for(int i=1;i<=n-1;i++)        {            for(int j=i+1;j<=n;j++)            {                q[cnt].i=i,q[cnt].j=j,q[cnt].sum=a[i]+a[j];                cnt++;            }        }        int maxn=0;        /*for(int i=1;i<=n;i++)        {            for(int j=0;j<cnt;j++)            {                if(i!=q[j].i&&i!=q[j].j)                    maxn=max(maxn,q[j].sum^a[i]);            }        }*/        for(int i=0;i<cnt;i++)        {            for(int j=1;j<=n;j++)            {                if(j!=q[i].i&&j!=q[i].j)                    maxn=max(maxn,q[i].sum^a[j]);            }        }        printf("%d\n",maxn);    }    return 0;}

PS：暴力的代码可以勉强A，而注释掉的循环会比AC代码多500-700ms（亲测），甚至有时不能A。但不知道为什么会多这些时间...