题目描述
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?
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
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e3+5;
int ch[32 * maxn][2], val[32 * maxn], sz;
void init() {
ch[0][0] = ch[0][1] = 0;
memset(val, 0, sizeof(val));
sz = 1;
}
void _insert(ll a) {
int u = 0;
for (int i = 32; i >= 0; i--) {
int c = ((a >> i) & 1);
if (!ch[u][c]) {
ch[sz][0] = ch[sz][1] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
val[u]++;
}
}
void _delete(ll a) {
int u = 0;
for (int i = 32; i >= 0; i--) {
int c = ((a>>i)&1);
u = ch[u][c];
val[u]--;
}
}
ll query(ll a) {
int u = 0;
ll ans = 0;
for (int i = 32; i >= 0; i--) {
int c = ((a>>i)&1);
if (c == 1) {
if (ch[u][0] && val[ch[u][0]]) {
ans += 1ll<<i;
u = ch[u][0];
} else {
u = ch[u][1];
}
} else {
if (ch[u][1] && val[ch[u][1]]) {
ans += 1ll<<i;
u = ch[u][1];
} else {
u = ch[u][0];
}
}
}
return ans;
}
ll a[1005];
int main() {
int T, n;
scanf("%d", &T);
while (T--) {
init();
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%lld", &a[i]);
sort(a, a + n);
for (int i = 0; i < n; i++)
_insert(a[i]);
ll maxx = 0;
for (int i = 0; i < n; i++) {
_delete(a[i]);
for (int j = i + 1; j < n; j++) {
_delete(a[j]);
ll ans = query(a[i] + a[j]);
maxx = max(maxx, ans);
_insert(a[j]);
}
_insert(a[i]);
}
printf("%lld\n", maxx);
}
return 0;
}