Team Formation

 

Time Limit:3000MS     Memory Limit:131072KB     64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3870

Description

For an upcoming programming contest, Edward, the headmaster of Marjar University, is forming a two-man team from N students of his university.

Edward knows the skill level of each student. He has found that if two students with skill level A and B form a team, the skill level of the team will be A ⊕ B, where ⊕ means bitwise exclusive or. A team will play well if and only if the skill level of the team is greater than the skill level of each team member (i.e. A ⊕ B > max{A, B}).

Edward wants to form a team that will play well in the contest. Please tell him the possible number of such teams. Two teams are considered different if there is at least one different team member.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (2 <= N <= 100000), which indicates the number of student. The next line contains N positive integers separated by spaces. The i^th integer denotes the skill level of i^th student. Every integer will not exceed 10⁹.

Output

For each case, print the answer in one line.

Sample Input

2
3
1 2 3
5
1 2 3 4 5

Sample Output

1
6

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MaxN = 1e5 + 10;
int a[MaxN], bit[50]; // bit[i]表示有多少个数的最高位的1在第i位上

void solve(int x) {
    int l = 31;
    while(l >= 0) {
        if(x & (1<<l)) {
            bit[l]++;
            return ;
        }
        l--;
    }
    return ;
}

int main() {
    int T, n;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        memset(bit, 0, sizeof(bit));
        for(int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            solve(a[i]);
        }
        int ans = 0;
        for(int i = 0; i < n; i++) {
            int l = 31;
            while(l >= 0) {
                if(a[i] & (1<<l)) break;
                l--;
            }
            while(l >= 0) {
                if(!(a[i] & (1<<l))) ans += bit[l];
                l--;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}