[51nod 2493] sum of binary distances [bit operation]

$link$

analysis ：

$n^2$ Violence is more You can count two $xor$ See how many there are $1$
Complexity $O(n^{2}logn)$

You can also take apart each number binary If this bit has $cn{t}_1$ individual $1$$cn{t}_0$ individual $0$ In the end $cn{t}_0\times cn{t}_1$ Different
Complexity $O(nlogn)$

CODE：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define reg register
using namespace std;
typedef long long ll;
const int N=1e4+5;
int n,a[N],ans,cnt0[N],cnt1[N];
void calc(int x)
{
    
	for(int i=0;i<=30;i++)
		((x>>i)&1)?cnt1[i]++:cnt0[i]++;
}
int main(){
    
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
    
		scanf("%d",&a[i]);
		calc(a[i]);
	}
	for(int i=0;i<=30;i++)
		ans+=cnt1[i]*cnt0[i];
	printf("%d",ans);
	return 0;
}