Codeforces Round #167 (Div. 2) B. Dima and Sequence（暴力）

B. Dima and Sequence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Dima got into number sequences. Now he's got sequence a₁, a₂, ..., a_n, consisting of n positive integers. Also, Dima has got a function f(x), which can be defined with the following recurrence:

• f(0) = 0;
• f(2·x) = f(x);
• f(2·x + 1) = f(x) + 1.

Dima wonders, how many pairs of indexes (i, j) (1 ≤ i < j ≤ n) are there, such that f(a_i) = f(a_j). Help him, count the number of such pairs.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵). The second line contains n positive integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

The numbers in the lines are separated by single spaces.

Output

In a single line print the answer to the problem.

Please, don't use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Sample test(s)

Input

3
1 2 4

Output

3

Input

3
5 3 1

Output

1

Note

In the first sample any pair (i, j) will do, so the answer is 3.

In the second sample only pair (1, 2) will do.

一直觉得会超时，谁知一次可以AC，看来还是不会分析复杂度。

#include <iostream>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define cti const int
#define dg(i) cout << "*" << i << endl;

int f[1000001];

void Init_f()  //先计算一百万以内的f（x），以减少Find函数的递归深度
{
    f[0] = 0;
    for(int i = 1; i < 1000001; i++)
        f[i] = f[i/2] + (i & 1);
}

int Find(int s)
{
    if(s < 1000001) return f[s];
    return Find(s / 2) + (s & 1);
}

int main()
{
    int n;
    LL ans;
    Init_f();
    while(scanf("%d", &n) != EOF)
    {
        map<int, int> M;
        int num;
        ans = 0;
        while(n--)
        {
            scanf("%d", &num);
            ans = ans + M[Find(num)]++;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}