ZOJ 3870



Team Formation

---

Time Limit: 3 Seconds      Memory Limit: 131072 KB

---

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

Edward knows the skill level of each student. He has found that if two students with skill levelA 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 containsN positive integers separated by spaces. The i^th integer denotes the skill level ofi^th student. Every integer will not exceed 10⁹.

Output

For each case, print the answer in one line.

Sample Input

231 2 351 2 3 4 5

Sample Output

16

题意：一组数里面选出两个数，使得这两个数异或之后大于它们中最大的数。问有多少个数。

思路：可以将所有数的最高位的1存进bit数组里面。后面只要满足另外一个数与其他数最高位那位对应的位为0的话，则异或后的数必然大于它们中较大的数。

与的运算：  1&1=1   1&0=0    0&1=0    异或运算   1^0=0    1^0=1   0^1=1   

所以只要判断 0&1==0 ---->   0^1=1  这样就将 0变成了1，所以异或后值自然变大。

关键点，预存储所有数最高位的1.

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;int a[100005];int bit[50];void solve(int x)   //存储所有数最高位的一{    int Bit=31;    while(Bit>=0)    {        if(x &(1<<Bit))        {            bit[Bit]++;               return;        }        Bit--;    }    return;}int main(){    int t;    cin>>t;    while(t--)    {        int n,i;        cin>>n;        ll ans=0;        memset(bit,0,sizeof(bit));        memset(a,0,sizeof(a));        for(i=0;i<n;i++)        {            cin>>a[i];            solve(a[i]);        }        for(i=0;i<n;i++)        {            int Bit=31;            while(Bit>=0)   //找到一个大的数高位的一            {                if(a[i]&(1<<Bit))                    break;                Bit--;              }            while(Bit>=0)  //之后只要将后面的0变成1，则必然数会变大。            {                if((a[i]&(1<<Bit))==0)                    ans+=bit[Bit];                Bit--;            }        }        cout<<ans<<endl;    }}