Codeforces Round #672 (Div. 2) B. Rock and Lever(位运算、思维）_综合

题意：给n个数， $a_i...a_n)$ 寻找一对 $a_i,a_j)$ 使得 $a_i 与 a_j>=a_i\oplus a_j$
题解：进行一些运算可知，当 $a_{i}$ 的最高位的1与 $a_j$ 的最高位的1位置相同时满足条件

#include<bits/stdc++.h>#pragma GCC optimize(2)
#define rep(i,a,n) for (int i=a;i<=n;i++)//i为循环变量，a为初始值，n为界限值，递增
#define per(i,a,n) for (int i=a;i>=n;i--)//i为循环变量， a为初始值，n为界限值，递减。using namespace std;
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0)
#define ll long long
#define ull unsigned long long
#define ld long double
const int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<60;
const int mod=1e9+7;ll read()//快读
{
    ll w = 1, s = 0;char ch = getchar();while (ch < '0' || ch>'9') {
     if (ch == '-') w = -1; ch = getchar(); }while (ch >= '0' && ch <= '9') {
     s = s * 10 + ch - '0';    ch = getchar(); }return s * w;
}
//以上个人模板----------------------------------------------------------
void solve(){
    int n,a[100010],cnt[50];ll ans=0;cin>>n;rep(i,1,n) cin>>a[i];memset(cnt,0,sizeof(cnt));rep(i,1,n){
    int tem=1,pos=1;while(tem*2<=a[i]){
    tem*=2;pos++;}ans+=cnt[pos];cnt[pos]++;}cout<<ans<<endl;
}int main(){
    int _;cin>>_;while(_--){
    solve();}return 0;
}