zoj 3870 Team Formation（位运算,超时）

For an upcoming programming contest, Edward, the headmaster of Marjar University, is forming a two-man team from Nstudents 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

231 2 351 2 3 4 5

Sample Output

16

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

#include<cstdio>  #include<algorithm>  #include<cstring>  #include<iostream>  #include<cmath>  #define N 1001010  #define ll long long    using namespace std;   int n,l;int a[N];  int num[N];  int p[33];  int b[33];      void init() {      p[0]=1;      for(int i=1; i<=30; i++)          p[i]=p[i-1]*2;  }    void change(int x) {      l=-1;      while(x) {          b[++l]=x%2;          x/=2;      }  }  int main() {      int t;      cin>>t;      init();      while(t--) {          cin>>n;          memset(num,0,sizeof num);          for(int i=0; i<n; i++) {              scanf("%d",&a[i]);              for(int j=30; j>=0; j--) {//求位数（第一位必是1）                 if(a[i]>=p[j]) {                      num[j]++;                      break;                  }              }          }          ll s=0;          for(int i=0; i<n; i++) {              change(a[i]);              for(int j=0; j<=l; j++) {                  if(b[j]==0) {  //这一位是0，另一个数的这一位是首位(1)                    s+=num[j];                  }              }          }          printf("%lld\n",s);      }  }