Team FormationTime Limit:3 Seconds Memory Limit:131072 KB
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 Output16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（i不是y的最高位所在位置），那么z=x^y之后，z > max(x, y)。#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\n", ans);}return 0;}
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