给定一个大小为n的数组arr []，我们需要找到所有子集的值之和。
先决条件：给定集合的子集总和
例子 ：
Input : arr[] = {1, 2, 3}
Output : 18
Total Subsets = 23 -1= 7
1 = 1
2 = 2
3 = 3
1 | 2 = 3
1 | 3 = 3
2 | 3 = 3
1 | 2 | 3 = 3
0(empty subset)
Now SUM of all these ORs = 1 + 2 + 3 + 3 +
3 + 3 + 3
= 18
Input : arr[] = {1, 2, 3}
Output : 18
天真的方法是对array []元素的所有可能组合进行“或”运算，然后对所有值进行求和。这种方法的时间复杂度呈指数增长，因此对于较大的n值来说，这不会更好。
一种有效的方法是找到有关OR属性的模式。现在再次考虑二进制形式的子集，例如：
1 = 001
2 = 010
3 = 011
1 | 2 = 011
1 | 3 = 011
2 | 3 = 011
1|2|3 = 011为了对数组中所有可能的元素取或，我们在这里考虑了第i位为1的所有可能子集。
现在，考虑所有结果OR的第i位，仅当子集中元素的所有第i位为0时，它才为零。
第i位为1的子集数=可能的子集总数–第i位为全0的子集。这里，总子集= 2 ^ n – 1和所有第i位为0的子集= 2 ^(所有元素的第i位为零的计数)数组的总和)– 1.现在，第i位为1的总子集OR =(2 ^ n-1)-(2 ^(第i位为零的计数)-1)。这些值为1的位贡献的总价值=总子集或第i个位为1 *(2 ^ i)。
现在，总和=(第i位为1的总子集)* 2 ^ i +(第i + 1位为i + 1的总子集)* 2 ^(i + 1)+………+(32位为1的总子集)* 2 ^ 32。
C++
// CPP code to find the OR_SUM
#include
using namespace std;
#define INT_SIZE 32
// function to find the OR_SUM
int ORsum(int arr[], int n)
{
// create an array of size 32
// and store the sum of bits
// with value 0 at every index.
int zerocnt[INT_SIZE] = { 0 };
for (int i = 0; i < INT_SIZE; i++)
for (int j = 0; j < n; j++)
if (!(arr[j] & 1 << i))
zerocnt[i] += 1;
// for each index the OR sum contributed
// by that bit of subset will be 2^(bit index)
// now the OR of the bits is 0 only if
// all the ith bit of the elements in subset
// is 0.
int ans = 0;
for (int i = 0; i < INT_SIZE; i++)
{
ans += ((pow(2, n) - 1) -
(pow(2, zerocnt[i]) - 1)) *
pow(2, i);
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3 };
int size = sizeof(arr) / sizeof(arr[0]);
cout << ORsum(arr, size);
return 0;
} Java
// Java code to find
// the OR_SUM
import java.io.*;
class GFG {
static int INT_SIZE = 32;
// function to find
// the OR_SUM
static int ORsum(int []arr, int n)
{
// create an array of size 32
// and store the sum of bits
// with value 0 at every index.
int zerocnt[] = new int[INT_SIZE] ;
for (int i = 0; i < INT_SIZE; i++)
for (int j = 0; j < n; j++)
if ((arr[j] & 1 << i) == 0)
zerocnt[i] += 1;
// for each index the OR
// sum contributed by that
// bit of subset will be
// 2^(bit index) now the OR
// of the bits is 0 only if
// all the ith bit of the
// elements in subset is 0.
int ans = 0;
for (int i = 0; i < INT_SIZE; i++)
{
ans += ((Math.pow(2, n) - 1) -
(Math.pow(2, zerocnt[i]) - 1)) *
Math.pow(2, i);
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3 };
int size = arr.length;
System.out.println(ORsum(arr, size));
}
}
// This code is contributed by Sam007Python3
INT_SIZE = 32
# function to find the OR_SUM
def ORsum(arr, n):
# create an array of size 32
# and store the sum of bits
# with value 0 at every index.
zerocnt = [0 for i in range(INT_SIZE)]
for i in range(INT_SIZE):
for j in range(n):
if not (arr[j] & (1 << i)):
zerocnt[i] += 1
# for each index the OR sum contributed
# by that bit of subset will be 2^(bit index)
# now the OR of the bits is 0 only if
# all the ith bit of the elements in subset
# is 0.
ans = 0
for i in range(INT_SIZE):
ans += ((2 ** n - 1) - (2 ** zerocnt[i] - 1)) * 2 ** i
return ans
# Driver code
if __name__ == "__main__":
arr= [1, 2, 3]
size = len(arr)
print(ORsum(arr, size))
# This code is contributed by vaibhav29498C#
// C# code to find
// the OR_SUM
using System;
class GFG {
static int INT_SIZE = 32;
// function to find
// the OR_SUM
static int ORsum(int []arr, int n)
{
// create an array of size 32
// and store the sum of bits
// with value 0 at every index.
int []zerocnt = new int[INT_SIZE] ;
for (int i = 0; i < INT_SIZE; i++)
for (int j = 0; j < n; j++)
if ((arr[j] & 1 << i) == 0)
zerocnt[i] += 1;
// for each index the OR
// sum contributed by that
// bit of subset will be
// 2^(bit index) now the OR
// of the bits is 0 only if
// all the ith bit of the
// elements in subset is 0.
int ans = 0;
for (int i = 0; i < INT_SIZE; i++)
{
ans += (int)(((Math.Pow(2, n) - 1) -
(Math.Pow(2, zerocnt[i]) - 1)) *
Math.Pow(2, i));
}
return ans;
}
// Driver Code
public static void Main()
{
int []arr = {1, 2, 3};
int size = arr.Length;
Console.Write(ORsum(arr, size));
}
}
// This code is contributed by nitin mittalPHP
Javascript
输出：
18时间复杂度： O(n)
辅助空间： O(n)