font ava display algo acm getchar 暴力 width miss

Chip Factory

Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 4414 Accepted Submission(s): 1954

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)⊕s

k
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 2 3 1 2 3 3 100 200 300

Sample Output 6 400

Source 2015ACM/ICPC亞洲區長春站-重現賽（感謝東北師大）

Recommend hujie | We have carefully selected several similar problems for you: 6263 6262 6261 6260 6259 讀不懂英語好吃虧啊。 一開始一直在想前面的$s_i+s_j$怎麽搞。 後來看了一下輸入，這麽不就是個逗比題麽。。 $s_i+s_j$只要暴力枚舉就好，建一顆01Trie樹，查詢最大的時候優先走不一樣的 查詢之前先把$s_i,s_j$刪除 查完之後再加進去

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN=1e6+10;
inline int read()
{
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}
#define debug(x) printf("%d",x);
struct node
{
    int val,ch[2];
    node(){val=ch[0]=ch[1]=0;}
    void clear(){val=ch[0]=ch[1]=0;}
}T[MAXN];
int a[MAXN],root=0,tot=0;
void Insert(int v)
{
    int now=root;
    for(int i=31;i>=0;i--)
    {
        int opt=(v&(1<<i))?1:0;
        if(!T[now].ch[opt]) 
            T[now].ch[opt]=++tot;
        now=T[now].ch[opt];
        T[now].val++;
     }
}
void Delet(int v)
{
    int now=root;
    for(int i=31;i>=0;i--)
    {
        int opt=(v&(1<<i))?1:0;
        now=T[now].ch[opt];
        T[now].val--;
     }
}
int Query(int v)
{
    int ans=0,now=root;
    for(int i=31;i>=0;i--)
    {
        int opt=(v&(1<<i))?1:0;
        if(T[T[now].ch[opt^1]].val) 
            ans+=1<<i,now=T[now].ch[opt^1];
        else 
            now=T[now].ch[opt];
    }
    return ans;
}
int main()
{
    freopen("a.in","r",stdin);
    int Test=read();
    while(Test--)
    {
        int N=read();
        for(int i=1;i<=N;i++) a[i]=read();
        for(int i=1;i<=4*N;i++) 
            T[i].clear();
        for(int i=1;i<=N;i++) 
            Insert(a[i]);
        int ans=0;
        for(int i=1;i<=N;i++)
        {
            for(int j=1;j<i;j++)
            {
                Delet(a[i]);Delet(a[j]);
                ans=max(ans,Query(a[i]+a[j]));
                Insert(a[i]);Insert(a[j]);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

HDU 5536 Chip Factory