完美和问题

给定一个整数数组arr[]和一个整数K ，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：想法是使用幂集概念找出所有子集。对于每个集合，检查集合的总和是否等于 K。如果相等，则打印该集合。
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_R

C#

// 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

Javascript

 
// JavaScript implementation of the above approach
 
 
// Function to print the subsets whose
// sum is equal to the given target K
function sumSubsets(set, n, target) {
    // Create the new array with length
    // equal to array set[] to create
    // binary array as per n(decimal number)
    let x = new Array(set.length);
    let j = set.length - 1;
 
    // Convert the array into binary array
    while (n > 0) {
        x[j] = n % 2;
        n = Math.floor(n / 2);
        j--;
    }
 
    let sum = 0;
 
    // Calculate the sum of this subset
    for (let 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) {
        document.write("{");
        for (let i = 0; i < set.length; i++)
            if (x[i] == 1)
                document.write(set[i] + ", ");
        document.write("}, ");
    }
}
 
// Function to find the subsets with sum K
function findSubsets(arr, K) {
    // Calculate the total no. of subsets
    let x = Math.pow(2, arr.length);
 
    // Run loop till total no. of subsets
    // and call the function for each subset
    for (let i = 1; i < x; i++)
        sumSubsets(arr, i, K);
}
 
// Driver code
 
let arr = [5, 10, 12, 13, 15, 18];
let K = 30;
findSubsets(arr, K);
 
 
// This code is contributed by gfgking
 

输出：

{12, 18, }, {5, 12, 13, }, {5, 10, 15, },

时间复杂度： 2 ^N
有效的方法：
这个问题也可以用动态规划解决。参考这篇文章。

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