zoj 3870 Team Formation（异或运算）

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5518

Team Formation 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

题意：给出n个数，每次选2个数，问一共有多少种选法使得选出的这两个数异或后的值，这两个数中的最大值还要大。

分析：异或运算：1^1=0， 1^0=1, 0^1=1, 0^0=0。

对于一个数，如果我们把x的二进制表示中最高位的0变成1，0前面的都不变，那么得到的这个新值肯定比x大。即：如果x的第i位为1（i为x的最高位的1所在位置），y的第i位为0，那么z=x^y之后，z > max(x, y)。

///#pragma comment (linker, "/STACK:102400000,102400000")

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <stack>
#include <vector>
#include <map>

using namespace std;

#define N 301000
#define INF 0xfffffff
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))

int main ()
{
    int t, n, a[N],b[N];
    scanf ("%d", &t);
    while (t--)
    {
        scanf ("%d", &n);
        met (b, 0);

        for (int i=1; i<=n; i++)
        {
            scanf ("%d", &a[i]);
            int tt = 31;
            while (tt >= 0)
            {
                if (a[i] & (1<<tt))
                {
                    b[tt]++;
                    break;
                }
                tt--;
            }
        }

        int ans = 0;
        for (int i=1; i<=n; i++)
        {
            int tt = 31;
            while (tt >= 0)
            {
                if (a[i] & (1<<tt))
                    break;
                tt--;
            }
            while (tt >= 0)
            {
                if (!(a[i] & (1<<tt)))
                    ans += b[tt];
                tt--;
            }
        }
        printf ("%d\n", ans);
    }
    return 0;
}