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:
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?
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
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
The first line of each test case is an integer n
1≤T≤1000
3≤n≤1000
0≤s
There are at most 10
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
Recommend
读不懂英语好吃亏啊。
一开始一直在想前面的$s_i+s_j$怎么搞。
后来看了一下输入，这么不就是个逗比题么。。
$s_i+s_j$只要暴力枚举就好，建一颗01Trie树，查询最大的时候优先走不一样的
查询之前先把$s_i,s_j$删除
查完之后再加进去
#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; }
