Chip Factory（暴力）_f - chip factory博客-优快云博客

本文链接：https://blog.youkuaiyun.com/stdio_xuege/article/details/82227866

博客围绕CPU芯片工厂的芯片序列号校验和计算问题展开。介绍了问题背景，给出输入输出格式及相关限制条件。作者看到题目通过率高，采用暴力方法解题并通过。涉及芯片序列号、校验和计算等信息技术相关内容。

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题目：

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 nn chips today, the ii-th chip produced this day has a serial number s_isi.

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:

\displaystyle max_{i, j, k}(s_i + s_j) \oplus s_kmaxi,j,k(si+sj)⊕sk

which i,j,ki,j,k are three different integers between 11 and nn. And \oplus⊕ is symbol of bitwise XOR.

Can you help John calculate the checksum number of today?

Input Format

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

The first line of each test case is an integer nn, indicating the number of chips produced today.

The next line has nn integers s_1,s_2,..,s_ns1,s2,..,sn, separated with single space, indicating serial number of each chip.

• 1\le T \le 10001≤T≤1000

• 3\le n\le 10003≤n≤1000

• 0\le s_i \le 10^90≤si≤109

• There are at most 1010 testcases with n > 100n>100

Output Format

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

样例输入复制

2
3
1 2 3
3
100 200 300

样例输出复制

6
400

分析：

看到这道题的通过率这么高，我就想用简单的方法————暴力试一下，结果就过了；

代码：

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int a[1005];
int main()
{
    int t,n,i,j,k,maxx;
    scanf("%d",&t);
    while(t--)
    {
        maxx=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(k=1;k<=n;k++)
            for(i=k+1;i<=n;i++)
                for(j=i+1;j<=n;j++)
                   {
                      maxx=max(maxx,(a[k]+a[j])^a[i]);
                      maxx=max(maxx,(a[i]+a[j])^a[k]);
                      maxx=max(maxx,(a[i]+a[k])^a[j]);
                   }
        printf("%d\n",maxx);
    }
    return 0;
}