完美和问题（打印给定和的所有子集）

给定一个整数数组和一个总和，任务是打印给定数组的所有子集，其总和等于给定的总和。
例子：

Input : arr[] = {2, 3, 5, 6, 8, 10}
        sum = 10
Output : 5 2 3
         2 8
         10

Input : arr[] = {1, 2, 3, 4, 5}
        sum = 10
Output : 4 3 2 1 
         5 3 2 
         5 4 1 

在 Tesco 询问

这个问题主要是子集和问题的扩展。这里我们不仅需要查找是否存在具有给定和的子集，还需要打印具有给定和的所有子集。
和上一篇文章一样，我们构建了一个二维数组 dp[][] ，如果从 0 到 i 的数组元素可以求和 j ，则 dp[i][j] 存储真。
填充dp[][]后，我们从dp[n-1][sum]开始递归遍历。对于被遍历的单元格，我们在到达它之前存储路径并考虑元素的两种可能性。
1) 元素包含在当前路径中。
2) 元素不包含在当前路径中。
每当总和变为 0 时，我们停止递归调用并打印当前路径。
下面是上述想法的一个实现。

C++

// C++ program to count all subsets with
// given sum.
#include 
using namespace std;
 
// dp[i][j] is going to store true if sum j is
// possible with array elements from 0 to i.
bool** dp;
 
void display(const vector& v)
{
    for (int i = 0; i < v.size(); ++i)
        printf("%d ", v[i]);
    printf("\n");
}
 
// A recursive function to print all subsets with the
// help of dp[][]. Vector p[] stores current subset.
void printSubsetsRec(int arr[], int i, int sum, vector& p)
{
    // If we reached end and sum is non-zero. We print
    // p[] only if arr[0] is equal to sun OR dp[0][sum]
    // is true.
    if (i == 0 && sum != 0 && dp[0][sum])
    {
        p.push_back(arr[i]);
        // Display Only when Sum of elements of p is equal to sum
          if (arr[i] == sum)
              display(p);
        return;
    }
 
    // If sum becomes 0
    if (i == 0 && sum == 0)
    {
        display(p);
        return;
    }
 
    // If given sum can be achieved after ignoring
    // current element.
    if (dp[i-1][sum])
    {
        // Create a new vector to store path
        vector b = p;
        printSubsetsRec(arr, i-1, sum, b);
    }
 
    // If given sum can be achieved after considering
    // current element.
    if (sum >= arr[i] && dp[i-1][sum-arr[i]])
    {
        p.push_back(arr[i]);
        printSubsetsRec(arr, i-1, sum-arr[i], p);
    }
}
 
// Prints all subsets of arr[0..n-1] with sum 0.
void printAllSubsets(int arr[], int n, int sum)
{
    if (n == 0 || sum < 0)
       return;
 
    // Sum 0 can always be achieved with 0 elements
    dp = new bool*[n];
    for (int i=0; i p;
    printSubsetsRec(arr, n-1, sum, p);
}
 
// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    int sum = 10;
    printAllSubsets(arr, n, sum);
    return 0;
}

Java

// A Java program to count all subsets with given sum.
import java.util.ArrayList;
public class SubSet_sum_problem
{
    // dp[i][j] is going to store true if sum j is
    // possible with array elements from 0 to i.
    static boolean[][] dp;
      
    static void display(ArrayList v)
    {
       System.out.println(v);
    }
      
    // A recursive function to print all subsets with the
    // help of dp[][]. Vector p[] stores current subset.
    static void printSubsetsRec(int arr[], int i, int sum,
                                         ArrayList p)
    {
        // If we reached end and sum is non-zero. We print
        // p[] only if arr[0] is equal to sun OR dp[0][sum]
        // is true.
        if (i == 0 && sum != 0 && dp[0][sum])
        {
            p.add(arr[i]);
            display(p);
            p.clear();
            return;
        }
      
        // If sum becomes 0
        if (i == 0 && sum == 0)
        {
            display(p);
            p.clear();
            return;
        }
      
        // If given sum can be achieved after ignoring
        // current element.
        if (dp[i-1][sum])
        {
            // Create a new vector to store path
            ArrayList b = new ArrayList<>();
            b.addAll(p);
            printSubsetsRec(arr, i-1, sum, b);
        }
      
        // If given sum can be achieved after considering
        // current element.
        if (sum >= arr[i] && dp[i-1][sum-arr[i]])
        {
            p.add(arr[i]);
            printSubsetsRec(arr, i-1, sum-arr[i], p);
        }
    }
      
    // Prints all subsets of arr[0..n-1] with sum 0.
    static void printAllSubsets(int arr[], int n, int sum)
    {
        if (n == 0 || sum < 0)
           return;
      
        // Sum 0 can always be achieved with 0 elements
        dp = new boolean[n][sum + 1];
        for (int i=0; i p = new ArrayList<>();
        printSubsetsRec(arr, n-1, sum, p);
    }
     
    //Driver Program to test above functions
    public static void main(String args[])
    {
        int arr[] = {1, 2, 3, 4, 5};
        int n = arr.length;
        int sum = 10;
        printAllSubsets(arr, n, sum);
    }
}
//This code is contributed by Sumit Ghosh

输出：

4 3 2 1
5 3 2
5 4 1

另一种方法：
对于数组中的每个元素，首先决定是否将其放入子集中。定义一个函数来处理所有这些。该函数在主函数被调用一次。声明了静态类字段，这些字段将由我们的函数操作。每次调用时，该函数检查字段的条件。在我们的例子中，它检查当前的总和是否等于给定的总和，并相应地增加相应的类字段。如果不是，它通过处理所有情况进行函数调用。所以函数调用的数量将等于案例的数量。因此，这里进行了两次调用——一次通过获取子集中的元素并增加当前总和，另一次通过不获取该元素。
下面是实现：

Java

// Java code to find the number of
// possible subset with given sum
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG {
     
    static int count;
    static int sum;
    static int n;
     
    // Driver code
    public static void main (String[] args) {
        count = 0;
        n = 5;
        sum = 10;
 
        int[] pat = {2, 3, 5, 6, 8, 10};
         
        f(pat, 0, 0);
         
        System.out.println(count);
    }
     
    // Function to select or not the array element
    // to form a subset with given sum
    static void f(int[] pat, int i, int currSum) {
        if (currSum == sum) {
            count++;
            return;
        }
        if (currSum < sum && i < n) {
            f(pat, i+1, currSum + pat[i]);
            f(pat, i+1, currSum);
        }
    }
}

输出 ：

4 3 2 1 
5 3 2 
5 4 1 

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