题目链接：?problemId=5518
For an upcoming programming contest, Edward, the headmaster of Marjar University, is forming a two-man team fromNstudents of his university.
Edward knows the skill level of each student. He has found that if two students with skill levelAandBform a team, the skill level of the team will beA⊕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 integerTindicating the number of test cases. For each test case:
The first line contains an integerN(2 <=N<= 100000), which indicates the number of student. The next line containsNpositive integers separated by spaces. Theithinteger denotes the skill level ofithstudent. Every integer will not exceed 109.
Output
For each case, print the answer in one line.
Sample Input231 2 351 2 3 4 5Sample Output16Author:LIN, XiSource:The 12th Zhejiang Provincial Collegiate Programming Contest
题意：给出 n 个数，每次选出两个数，问一共有多少种选法能使得选出的这两个数的异或值比这两个数中最大的那个数还大。也就是z = x^y之后，z > max(x, y)。PS:对于一个数，如果我们把 x 的二进制表示中的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。代码如下：#include <cstdio>#include <cstring>#include <cmath>#define maxn 100017int a[maxn], c[maxn];void cal(int m){int l = 31;while(l >= 0){if( m & (1<<l)){c[l]++;return ;}l–;}return ;}int main(){int t;int n;scanf("%d",&t);while(t–){memset(c,0,sizeof(c));scanf("%d",&n);for(int i = 0; i < n; i++){scanf("%d",&a[i]);cal(a[i]);}int ans = 0;for(int i = 0; i < n; i++){int l = 31;while(l >= 0){if((1<<l) & a[i])//找到最前面那个一的位置{break;}l–;}while(l >= 0){if(!((1<<l) & a[i])){ans+=c[l];}l–;}}printf("%d\n",ans);}return 0;}
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