Dima and Sequence CodeForces - 272B （思维题）

Dima got into number sequences. Now he’s got sequence a1, a2, …, an, 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(ai) = f(aj). Help him, count the number of such pairs.

Input
The first line contains integer n (1 ≤ n ≤ 10^5). The second line contains n positive integers a1, a2, …, an (1 ≤ ai ≤ 10^9).

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.

Example
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.

大致题意：给你n个数，问你有多少对数i，j满足i< j且f（i）==f（j）。

思路：假设这n个数中有x个数对应的f（）是相同的，那么对于这x个数我们可以找出x*（x-1）/2对满足条件，所以问题就转化为求有多少数对应的f（）是相同的，最后将所有的答案相加即可。然后观察题目中所给的式子，不难发现，当x是偶数时，f（x）=f（x/2），值不会改变，当x是奇数时，f(x)=f((x-1)/2)+1，值会加一，所以我们可以从后往前推，当x是偶数时就除以2，值不变，当x是奇数时就减一除以2，值加一。因为x<=1e9,所以结果不会超过30。我们开个长度为30的数组记录下即可。

代码如下

#include
#include
#include
#include
#include
#define LL long long 
using namespace std; 
LL f[35];

int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int x;
        scanf("%d",&x);
        int sum=0;
        while(x)
        {
            if(x&1)
            {
                sum++;
                x=(x-1)/2;
            }
            else 
            x=x/2;
        }
        f[sum]++;
    }
    LL ans=0;
    for(int i=0;i<35;i++)
    ans+=f[i]*(f[i]-1)/2;
    printf("%I64d",ans);
    return 0; 
}