Codeforces Global Round 8 D - AND, OR and square sum 尽量往大的数字上移动

#include<map>
#include<queue>
#include<time.h>
#include<limits.h>
#include<cmath>
#include<ostream>
#include<iterator>
#include<set>
#include<stack>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep_1(i,m,n) for(int i=m;i<=n;i++)
#define mem(st) memset(st,0,sizeof st)
int read()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')             //判断正负
        flag=1;
    else if(ch>='0'&&ch<='9')           //得到完整的数
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef pair<double,double> pdd;
const int inf = 0x3f3f3f3f;
const int N = 1000100;
int n;
int a[N], cnt[N];
int main()
{
    //1具有传递性，那么就把1尽量往大的数字上传
    n = read();
    for (int i = 1; i <= n; i ++)
    {
        a[i] = read();
        for (int j = 0; j <= 20; j ++)
            if (a[i] >> j & 1) cnt[j] ++;
    }
    long long ans = 0;
    for (int i = 1, tar; i <= n; i ++)
    {
        tar = 0;
        for (int j = 0; j <= 20; j ++)
            if (cnt[j])
                tar |= (1 << j), cnt[j] --;
        ans += 1ll * tar * tar;
    }
    printf("%lld\n", ans);
    return 0;
}