给定一个包含N个整数的数组arr ，任务是计算具有相同设置位数目的元素对的可能数目。
例子：
Input: N = 8, arr[] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 9
Explanation:
Elements with 1 set bit: 1, 2, 4, 8
Elements with 2 set bits: 3, 5, 6
Elements with 3 set bits: 7
Hence, {1, 2}, {1, 4}, {1, 8}, {2, 4}, {2, 8}, {4, 8}, {3, 5}, {3, 6}, and {5, 6} are the possible such pairs.
Input: N = 12, arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Output: 22
方法：
- 将所有数字的设置位预计算并存储到bitscount []中，直到数组的最大元素。对于所有2的幂，将1存储在其相应的索引处。之后，通过以下关系计算其余元素的设置位数：
bitscount[i] = bitscount[previous power of 2] + bitscount[i – previous power of 2]
- 将设置位的频率存储在映射的数组元素中。
- 为每个设置的位计数添加可能的对数。如果X个元素的置位位数相同，则它们之间可能的对数为X *(X – 1)/ 2 。
- 打印此类对的总数。
下面的代码是上述方法的实现：
C++14
// C++ Program to count
// possible number of pairs
// of elements with same
// number of set bits.
#include
using namespace std;
// Function to return the
// count of Pairs
int countPairs(int arr[], int N)
{
// Get the maximum element
int maxm = *max_element(arr, arr + N);
int i, k;
// Array to store count of bits
// of all elements upto maxm
int bitscount[maxm + 1] = { 0 };
// Store the set bits
// for powers of 2
for (i = 1; i <= maxm; i *= 2)
bitscount[i] = 1;
// Compute the set bits for
// the remaining elements
for (i = 1; i <= maxm; i++) {
if (bitscount[i] == 1)
k = i;
if (bitscount[i] == 0) {
bitscount[i]
= bitscount[k]
+ bitscount[i - k];
}
}
// Store the frequency
// of respective counts
// of set bits
map setbits;
for (int i = 0; i < N; i++) {
setbits[bitscount[arr[i]]]++;
}
int ans = 0;
for (auto it : setbits) {
ans += it.second
* (it.second - 1) / 2;
}
return ans;
}
int main()
{
int N = 12;
int arr[] = { 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12 };
cout << countPairs(arr, N);
return 0;
} Java
// Java program to count possible
// number of pairs of elements
// with same number of set bits
import java.util.*;
import java.lang.*;
class GFG{
// Function to return the
// count of Pairs
static int countPairs(int []arr, int N)
{
// Get the maximum element
int maxm = arr[0];
for(int j = 1; j < N; j++)
{
if (maxm < arr[j])
{
maxm = arr[j];
}
}
int i, k = 0;
// Array to store count of bits
// of all elements upto maxm
int[] bitscount = new int[maxm + 1];
Arrays.fill(bitscount, 0);
// Store the set bits
// for powers of 2
for(i = 1; i <= maxm; i *= 2)
bitscount[i] = 1;
// Compute the set bits for
// the remaining elements
for(i = 1; i <= maxm; i++)
{
if (bitscount[i] == 1)
k = i;
if (bitscount[i] == 0)
{
bitscount[i] = bitscount[k] +
bitscount[i - k];
}
}
// Store the frequency
// of respective counts
// of set bits
Map setbits = new HashMap<>();
for(int j = 0; j < N; j++)
{
setbits.put(bitscount[arr[j]],
setbits.getOrDefault(
bitscount[arr[j]], 0) + 1);
}
int ans = 0;
for(int it : setbits.values())
{
ans += it * (it - 1) / 2;
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int N = 12;
int []arr = { 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12 };
System.out.println(countPairs(arr, N));
}
}
// This code is contributed by offbeat Python3
# Python3 program to count possible number
# of pairs of elements with same number
# of set bits.
# Function to return the
# count of Pairs
def countPairs(arr, N):
# Get the maximum element
maxm = max(arr)
i = 0
k = 0
# Array to store count of bits
# of all elements upto maxm
bitscount = [0 for i in range(maxm + 1)]
i = 1
# Store the set bits
# for powers of 2
while i <= maxm:
bitscount[i] = 1
i *= 2
# Compute the set bits for
# the remaining elements
for i in range(1, maxm + 1):
if (bitscount[i] == 1):
k = i
if (bitscount[i] == 0):
bitscount[i] = (bitscount[k] +
bitscount[i - k])
# Store the frequency
# of respective counts
# of set bits
setbits = dict()
for i in range(N):
if bitscount[arr[i]] in setbits:
setbits[bitscount[arr[i]]] += 1
else:
setbits[bitscount[arr[i]]] = 1
ans = 0
for it in setbits.values():
ans += it * (it - 1) // 2
return ans
# Driver Code
if __name__=='__main__':
N = 12
arr = [ 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12 ]
print(countPairs(arr, N))
# This code is contributed by pratham76C#
// C# program to count
// possible number of pairs
// of elements with same
// number of set bits.
using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
class GFG{
// Function to return the
// count of Pairs
static int countPairs(int []arr, int N)
{
// Get the maximum element
int maxm = -int.MaxValue;
for(int j = 0; j < N; j++)
{
if (maxm < arr[j])
{
maxm = arr[j];
}
}
int i, k = 0;
// Array to store count of bits
// of all elements upto maxm
int []bitscount = new int[maxm + 1];
Array.Fill(bitscount, 0);
// Store the set bits
// for powers of 2
for(i = 1; i <= maxm; i *= 2)
bitscount[i] = 1;
// Compute the set bits for
// the remaining elements
for(i = 1; i <= maxm; i++)
{
if (bitscount[i] == 1)
k = i;
if (bitscount[i] == 0)
{
bitscount[i] = bitscount[k] +
bitscount[i - k];
}
}
// Store the frequency
// of respective counts
// of set bits
Dictionary setbits = new Dictionary();
for(int j = 0; j < N; j++)
{
if (setbits.ContainsKey(bitscount[arr[j]]))
{
setbits[bitscount[arr[j]]]++;
}
else
{
setbits[bitscount[arr[j]]] = 1;
}
}
int ans = 0;
foreach(KeyValuePair it in setbits)
{
ans += it.Value * (it.Value - 1) / 2;
}
return ans;
}
// Driver Code
public static void Main(string[] args)
{
int N = 12;
int []arr = { 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12 };
Console.Write(countPairs(arr, N));
}
}
// This code is contributed by rutvik_56 输出：
22