ZOJ - 3870 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.

 

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.

题意：给出n个数，找出满足 a^b > max(a,b)的组合数。

分析：对于两个不相等的数给出两个数的二进制（右边为地位，右对齐）

1 1 0 1 0 0 1

0 0 1 0 0 1 0

经观察发现只需要满足较小的数的最高位为1，较大数的对应位为零，就会使数变大。

故只需要统计 cnt[k] 表示以k位为最高位的数的个数。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int N = 1E5+100;
int a[N];
int cnt[100],b[100];
int main(){
	int t;
	cin>>t;
	while(t--){
		int n,k;
		long long ans = 0;
		memset(cnt,0,sizeof cnt);
		scanf("%d",&n);
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);	
		}
		sort(a,a+n);
		for(int i=0;i<n;i++){
			k = 0;
			while(a[i]){
				b[k] = a[i]%2;
				if(b[k] == 0)ans += cnt[k];
				a[i] /= 2;	
				k++;
			}
			cnt[k-1]++;
		}
		printf("%lld\n",ans);
	}
	return 0;
}