标签：process tle include center others ref rod arch mes
Chip Factory
Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 5394 Accepted Submission(s): 2422
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
Sample Output
Source
时间充裕，暴力一发0.0
1 #include <iostream>
2 #include<cstring>
3 #include<algorithm>
4 #include<cstdio>
5 #include<cmath>
6 using namespace std;
7 int a[1005];
8 int main()
9 {
10 int t;
11 scanf("%d",&t);
12 while(t--)
13 {
14 int n,m=-999;
15 scanf("%d",&n);
16 for(int i=0;i<n;i++)
17 scanf("%d",&a[i]);
18 for(int i=0;i<n;i++){
19 for(int j=i+1;j<n;j++){
20 for(int k=j+1;k<n;k++){
21 m=max(m,(a[i]+a[j])^a[k]);
22 m=max(m,(a[j]+a[k])^a[i]);
23 m=max(m,(a[i]+a[k])^a[j]);
24 }
25 }
26 }
27 printf("%d\n",m);
28 }
29 return 0;
30 }
View Code
HDU 5536--Chip Factory(暴力)
标签：process tle include center others ref rod arch mes
原文地址：https://www.cnblogs.com/FlyerBird/p/9740342.html
