标签：-- tor scan tin dex 思路 const ons proc
原文链接：https://blog.csdn.net/qq_41021816/java/article/details/82934486
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 speci?cally, 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 de?ned 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?
输入
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
输出
For each test case, please output an integer indicating the checksum number in a line.
样例输入
2
3
1 2 3
3
100 200 300
样例输出
6
400
题目大意：
给 N 个数，在这 N 个数里找到三个值 i， j，k 使得（i+j）⊕ k 最大，输出这个最大值。
解题思路：
把每个数字看成一个0101字符串插入倒Trie树中去，枚举i和j，然后把si和sj从Trie树中删去。
然后在Trie树中贪心找到能与si+sj异或得到的最大值。
具体匹配的过程中是这样的，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。
另外删除操作是这样实现的，我们每个节点记录一个num值。
插入时对所有经过节点的num值加1，删除就将对应节点的num值减1。
在树上匹配的时候就只走那些num值为正的节点。
#include <bits/stdc++.h>
#define ll long long int
#define INF 0x3f3f3f3f
using namespace std;
const int MAXN = 1e3+10;
int ch[MAXN*32][2];
ll value[MAXN*32];
ll b[MAXN];
int num[MAXN*32];
int node_cnt;
void init()
{
memset(ch[0], 0, sizeof(ch[0]));
node_cnt = 1;
}
void Insert(ll x)
{
int cur = 0;
for(int i = 32; i >= 0; i--){
int index = (x>>i)&1;
if(!ch[cur][index]){
memset(ch[node_cnt], 0, sizeof(ch[node_cnt]));
ch[cur][index] = node_cnt;
value[node_cnt] = 0;
num[node_cnt++] = 0;
}
cur = ch[cur][index];
num[cur]++;
}
value[cur] = x;
}
void update(ll x, int d)
{
int cur = 0;
for(int i =32; i >= 0; i--){
int index = (x>>i)&1;
cur = ch[cur][index];
num[cur]+=d;
}
}
ll query(ll x)
{
int cur = 0;
for(int i = 32; i >= 0; i--)
{
int index = (x>>i)&1;
if(ch[cur][index^1] && num[ch[cur][index^1]]) cur = ch[cur][index^1];
else cur = ch[cur][index];
}
return x^value[cur];
}
int main()
{
int T_case, N;
scanf("%d", &T_case);
while(T_case--)
{
scanf("%d", &N);
init();
for(int i = 1; i <= N; i++)
{
scanf("%lld", &b[i]);
Insert(b[i]);
}
ll ans = 0;
for(int i = 1; i <= N; i++){
for(int j = 1; j <= N; j++){
if(i == j) continue;
update(b[i], -1);
update(b[j], -1);
ans = max(ans, query(b[i]+b[j]));
update(b[i], 1);
update(b[j], 1);
}
}
printf("%lld\n", ans);
}
return 0;
}
然而。。。。
#include <cstdio>
#include <algorithm>
using namespace std;
int a[1005];
int main()
{
int T;
scanf("%d", &T);
while(T --)
{
int n, ma = 0;
scanf("%d" ,&n);
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
for(int i = 0; i < n; i++)
for(int j = i + 1; j < n; j++)
for(int k = j + 1; k < n; k++)
{
ma = max(ma, (a[i] + a[j]) ^ a[k]);
ma = max(ma, (a[i] + a[k]) ^ a[j]);
ma = max(ma, (a[j] + a[k]) ^ a[i]);
}
printf("%d\n", ma);
}
}
标签：-- tor scan tin dex 思路 const ons proc
原文地址：https://www.cnblogs.com/cutemush/p/12623323.html
