HDU 5536 Chip Factory (暴力+技巧优化)

最新推荐文章于 2019-01-18 16:04:39 发布

最新推荐文章于 2019-01-18 16:04:39 发布 · 283 阅读

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#暴力

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本文介绍了一种CPU芯片工厂的校验计算方法，通过计算芯片串号的最大异或和来确保产品的一致性和准确性。针对大量芯片的管理难题，提出了一种简单而有效的暴力求解策略。

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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 sisi.

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:

max i, j, k (s i + s j) \oplus s k

which i,j,ki,j,k are three different integers between 11 and nn. 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 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 s1,s2,..,sns1,s2,..,sn, separated with single space, indicating serial number of each chip.

1≤T≤10001≤T≤1000
3≤n≤10003≤n≤1000
0≤si≤1090≤si≤109
There are at most 1010 testcases with n>100n>100

Output

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

Sample Input

2
3
1 2 3
3
100 200 300

Sample Output

6
400

【题解】

这题正解是字典树，不过根据给的时间看，可以暴力解，还是有那么一点点难度，直接暴力会T，稍微加一点优化，就过了，毕竟有9s时间，注意代码中红色部分，优化的地方。

【AC代码】

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string.h>
#include<math.h>
#include<stack>
using namespace std;
typedef long long ll;
const int inf=-1e8;
const int N=1005;
int m,n;
ll a[N];

bool cmp(int x,int y)
{
    return x>y;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&m);
        for(int i=1;i<=m;++i)
        {
            scanf("%lld",&a[i]);
        }
        sort(a+1,a+m+1,cmp);
        ll ans=-1;
        for(int i=1;i<=m;++i)
        {
            for(int j=i+1;j<=m;++j)
            {
                 for(int k=j+1;k<=m;++k)
                 {
                     ans = max(ans,(a[i]+a[j])^a[k]);
                     ans = max(ans,(a[i]+a[k])^a[j]);
                     ans = max(ans,(a[j]+a[k])^a[i]);
                 }
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}