CQUOJ 9884 Dima and Sequence

 

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 Input

Input

3
1 2 4

Output

3

Input

3
5 3 1

Output

1

Hint

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.

 

/*
2016年4月24日15:54:49

分析过程：
如果是偶数，那么直接等于除以2，如:  f(8) = f(4) = f(2) = f(1)
如果是奇数，那么直接等于除以2之后在加1，
    如：f(19) = f(9) + 1 = f(4) + 2 = f(2) + 2 = f(1) + 2 = f(0) + 3 = 3 
可以证明，左右数字都能经过不断地除以2，最后得到式子：f(0) + x = x 
第一、通过观察可知，只要是比1大的数，经过不断除以2，最后都能得到1，不管是奇数还是偶数
第二、通过公式可知，如果是偶数，就是像是8，它除以2之后，并不会增加x的值，
    只有奇数除以2之后，会把余出的1加在函数f外面的x上 

所以,问题的关键在于：在原来的除以直到除得0为止，一共有多少个商是奇数(包括原数在内）
然后得到一个值 k, num[k]   表示 所给的n个数中 值是k的个数 
最后只要看经过计算后的数列中，每个数字出现的次数，
通过排列组合(C n 2  你懂得...)，计算出有多少对相同的数对 

*/
# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <queue>
# include <vector>
# include <cmath>
# define LL long long 
# define INF 0x3f3f3f3f
using namespace std;
const int N = 1e5 + 5;
LL num[34];  
//  题目给的 a 最大不超过 1e9  即就算它二进制数上全是1  k值也不超过 32 
//   若 a = 63   二进制数为 0000111111   k = 6 ......

int main(void)
{
    int n, i, a;
    LL ans, k;
    while (~scanf("%d", &n))
    {
        memset(num, 0, sizeof(num));
        for (i = 1; i <= n; i++){
            scanf("%d", &a);
            k = 0;
            while (a){
                if (a % 2 == 1) k++;
                a >>= 1;
            }
            num[k]++;
        }
        ans = 0;
        for (i = 1; i < 34; i++){
            ans += num[i] * (num[i]-1) / 2;
        }
        printf("%I64d\n", ans);
    } 
    
    return 0;    
}