9264: Chip Factory

時間限制: 5 Sec  記憶體限制: 128 MB
提交: 268  解決: 61
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題目描述

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:

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?

輸入

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≤s i≤109
• There are at most 10 testcases with n > 100

輸出

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

樣例輸入

2
3
1 2 3
3
100 200 300

樣例輸出

6
400

來源/分類

題意:給出n個數,求其中不同的i,j,k使得(ai+aj)^ak的值最大,輸出最大值.

題解:第一次見到這種處理方式是在CF中,當時還感覺是個很騷的操作,現在發現這居然是基本操作.....

可以將每個數字的二進位制構造字典樹,那麼對於一個數查詢最大值,它的每一位二進位制就是一層,若它這一位是0,那麼你就要在字典樹的這一層查詢1,若這一位是1,那麼你就要查詢0(查詢到0這個數不會變化,查詢到1則會變大或變小),因為i,j,k不能相同,所以每次查詢列舉i,j就要在字典樹中把這兩個數字刪去,然後再新增上。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+7;
 
ll a[maxn];
int tire[maxn][2],tot,sz[maxn];
 
void ins(ll x)
{
    int rt=1;
    sz[rt]++;
    for(ll i=30;i>=0;i--)
    {
        int u=x&(1<<i)? 1:0;
        if(!tire[rt][u]) tire[rt][u]=++tot;
        rt=tire[rt][u];
        sz[rt]++;
    }
}
 
void del(ll x)
{
    int rt=1;
    sz[rt]--;
    for(ll i=30;i>=0;i--)
    {
        int u=x&(1<<i)? 1:0;
        rt=tire[rt][u];
        sz[rt]--;
    }
}
 
ll find(ll x)
{
    int rt=1;
    for(ll i=30;i>=0;i--)
    {
        int u=x&(1<<i)?1:0;
        if(u)
        {
            if( tire[rt][0] && sz[ tire[rt][0] ] )
                rt=tire[rt][0];
            else
                rt=tire[rt][1],x^=(1<<i);
        }
        else
        {
            if( tire[rt][1] && sz[ tire[rt][1] ] )
                rt=tire[rt][1],x^=(1<<i);
            else
                rt=tire[rt][0];
        }
    }
 
    return x;
}
 
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(tire,0,sizeof(tire));
        memset(sz,0,sizeof(sz));
        tot=1;
 
        int n;
        scanf("%d",&n);
 
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            ins(a[i]);
        }
 
        ll ans=0;
        for(int i=1;i<=n;i++)
        {
            del(a[i]);
            for(int j=i+1;j<=n;j++)
            {
                del(a[j]);
                ans=max(ans,find(a[i]+a[j]));
                ins(a[j]);
            }
            ins(a[i]);
        }
 
        printf("%lld\n",ans);
    }
 
    return 0;
}