hdu5536 Chip Factory

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)⊕sk

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

231 2 33100 200 300

 

Sample Output

6400

 

题意：给你n个数，让你找出3个不同下标对应的数，使得(s[i]+s[j])^s[k]的值最大。

思路：我们可以建一棵trie树，先把n个数都插入，然后枚举n*n个(i,j),先把trie树上的s[i],s[j]先消去，然后在树上找到异或和最大的k。我们每个节点记录一个val值。
插入时对所有经过节点的val值加1，删除就将对应节点的val值减1。在树上匹配的时候就只走那些val值为正的节点。匹配的过程中，首先看树中最高位能否异或得到1。
能的话就往能的那个方向走，否则往另外一个方向走。

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;typedef long long ll;#define inf 99999999#define pi acos(-1.0)#define maxnode 100000#define maxn 1005int a[maxn];int sz,tot;int ch[maxnode][2];int val[maxnode];char d[maxn][40];void init(){    sz=0;    memset(ch,0,sizeof(ch));    memset(val,0,sizeof(val));}int idx(char c){    return c-'0';}void update(char *s,int num){    int i,u=0,j;    int len=strlen(s);    char s1[40];    for(i=0;i<len;i++){        int c=idx(s[i]);        if(!ch[u][c]){            sz++;            val[sz]+=num;            ch[u][c]=sz;            u=sz;        }        else{            u=ch[u][c];            val[u]+=num;        }    }}int b[40],cnt;int chazhao(char *s){    int i,j,u=0;    int len=strlen(s);    int num=0;    for(i=0;i<len;i++){        int c=idx(s[i]);        if(val[ch[u][1^c ] ]){            b[i]=1;            u=ch[u][1^c ];        }        else{            b[i]=0;            u=ch[u][c];        }        num=num*2+b[i];    }    return num;}int main(){    int n,m,i,j,T,t,k;    char s[40];    scanf("%d",&T);    while(T--)    {        init();        memset(s,0,sizeof(s));        scanf("%d",&n);        for(i=1;i<=n;i++){            scanf("%d",&a[i]);            t=a[i];            int tot=0;            while(t){                s[tot++]=t%2+'0';                t>>=1;            }            while(tot<35){                s[tot++]='0';            }            s[tot]='\0';            reverse(s,s+tot);            update(s,1);            strcpy(d[i],s);        }        int ans=0;        for(i=1;i<n;i++){            update(d[i],-1);            for(j=i+1;j<=n;j++){                int num=a[i]+a[j];                update(d[j],-1);                tot=0;                char str[40];                while(num){                    str[tot++]=num%2+'0';                    num>>=1;                }                while(tot<35){                    str[tot++]='0';                }                str[tot]='\0';                reverse(str,str+tot);                cnt=0;                int temp=chazhao(str);                ans=max(ans,temp);                update(d[j],1);            }            update(d[i],1);        }        printf("%d\n",ans);    }    return 0;}