给定整数arr []和整数K的数组，任务是打印给定数组的所有子集，其总和等于给定目标K。
例子：
Input: arr[] = {5, 10, 12, 13, 15, 18}, K = 30
Output: {12, 18}, {5, 12, 13}, {5, 10, 15}
Explanation:
Subsets with sum 30 are:
12 + 18 = 30
5 + 12 + 13 = 30
5 + 10 + 15 = 30
Input: arr[] = {1, 2, 3, 4}, K = 5
Output: {2, 3}, {1, 4}
方法：想法是使用Power Set概念找出所有子集。对于每个集合，请检查集合的总和是否等于K。如果相等，则打印该组。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to print the subsets whose
// sum is equal to the given target K
void sumSubsets(vector set, int n, int target)
{
// Create the new array with size
// equal to array set[] to create
// binary array as per n(decimal number)
int x[set.size()];
int j = set.size() - 1;
// Convert the array into binary array
while (n > 0)
{
x[j] = n % 2;
n = n / 2;
j--;
}
int sum = 0;
// Calculate the sum of this subset
for (int i = 0; i < set.size(); i++)
if (x[i] == 1)
sum = sum + set[i];
// Check whether sum is equal to target
// if it is equal, then print the subset
if (sum == target)
{
cout<<("{");
for (int i = 0; i < set.size(); i++)
if (x[i] == 1)
cout << set[i] << ", ";
cout << ("}, ");
}
}
// Function to find the subsets with sum K
void findSubsets(vector arr, int K)
{
// Calculate the total no. of subsets
int x = pow(2, arr.size());
// Run loop till total no. of subsets
// and call the function for each subset
for (int i = 1; i < x; i++)
sumSubsets(arr, i, K);
}
// Driver code
int main()
{
vector arr = { 5, 10, 12, 13, 15, 18 };
int K = 30;
findSubsets(arr, K);
return 0;
}
// This code is contributed by mohit kumar 29 Java
// Java implementation of the above approach
import java.util.*;
class GFG {
// Function to print the subsets whose
// sum is equal to the given target K
public static void sumSubsets(
int set[], int n, int target)
{
// Create the new array with size
// equal to array set[] to create
// binary array as per n(decimal number)
int x[] = new int[set.length];
int j = set.length - 1;
// Convert the array into binary array
while (n > 0) {
x[j] = n % 2;
n = n / 2;
j--;
}
int sum = 0;
// Calculate the sum of this subset
for (int i = 0; i < set.length; i++)
if (x[i] == 1)
sum = sum + set[i];
// Check whether sum is equal to target
// if it is equal, then print the subset
if (sum == target) {
System.out.print("{");
for (int i = 0; i < set.length; i++)
if (x[i] == 1)
System.out.print(set[i] + ", ");
System.out.print("}, ");
}
}
// Function to find the subsets with sum K
public static void findSubsets(int[] arr, int K)
{
// Calculate the total no. of subsets
int x = (int)Math.pow(2, arr.length);
// Run loop till total no. of subsets
// and call the function for each subset
for (int i = 1; i < x; i++)
sumSubsets(arr, i, K);
}
// Driver code
public static void main(String args[])
{
int arr[] = { 5, 10, 12, 13, 15, 18 };
int K = 30;
findSubsets(arr, K);
}
}Python3
# Python3 implementation of the above approach
# Function to print the subsets whose
# sum is equal to the given target K
def sumSubsets(sets, n, target) :
# Create the new array with size
# equal to array set[] to create
# binary array as per n(decimal number)
x = [0]*len(sets);
j = len(sets) - 1;
# Convert the array into binary array
while (n > 0) :
x[j] = n % 2;
n = n // 2;
j -= 1;
sum = 0;
# Calculate the sum of this subset
for i in range(len(sets)) :
if (x[i] == 1) :
sum += sets[i];
# Check whether sum is equal to target
# if it is equal, then print the subset
if (sum == target) :
print("{",end="");
for i in range(len(sets)) :
if (x[i] == 1) :
print(sets[i],end= ", ");
print("}, ",end="");
# Function to find the subsets with sum K
def findSubsets(arr, K) :
# Calculate the total no. of subsets
x = pow(2, len(arr));
# Run loop till total no. of subsets
# and call the function for each subset
for i in range(1, x) :
sumSubsets(arr, i, K);
# Driver code
if __name__ == "__main__" :
arr = [ 5, 10, 12, 13, 15, 18 ];
K = 30;
findSubsets(arr, K);
# This code is contributed by Yash_RC#
// C# implementation of the above approach
using System;
class GFG
{
// Function to print the subsets whose
// sum is equal to the given target K
public static void sumSubsets(
int []set, int n, int target)
{
// Create the new array with size
// equal to array set[] to create
// binary array as per n(decimal number)
int []x = new int[set.Length];
int j = set.Length - 1;
// Convert the array into binary array
while (n > 0)
{
x[j] = n % 2;
n = n / 2;
j--;
}
int sum = 0;
// Calculate the sum of this subset
for (int i = 0; i < set.Length; i++)
if (x[i] == 1)
sum = sum + set[i];
// Check whether sum is equal to target
// if it is equal, then print the subset
if (sum == target)
{
Console.Write("{");
for (int i = 0; i < set.Length; i++)
if (x[i] == 1)
Console.Write(set[i] + ", ");
Console.Write("}, ");
}
}
// Function to find the subsets with sum K
public static void findSubsets(int[] arr, int K)
{
// Calculate the total no. of subsets
int x = (int)Math.Pow(2, arr.Length);
// Run loop till total no. of subsets
// and call the function for each subset
for (int i = 1; i < x; i++)
sumSubsets(arr, i, K);
}
// Driver code
public static void Main(String []args)
{
int []arr = { 5, 10, 12, 13, 15, 18 };
int K = 30;
findSubsets(arr, K);
}
}
// This code is contributed by 29AjayKumar输出：
{12, 18, }, {5, 12, 13, }, {5, 10, 15, },
时间复杂度： 2 N
高效方法：
使用动态编程也可以解决此问题。请参阅本文。