给定长度为N且数字为K的数组arr [] ，任务是查找元素和为K的数组的所有子序列
例子：
Input: arr[] = {1, 2, 3}, K = 3
Output:
1 2
3
Input: arr[] = {17, 18, 6, 11, 2, 4}, K = 6
Output:
2 4
6
方法：
这个想法是使用锯齿状的数组来存储不同长度的数组的子序列，对于数组中的每个元素，主要有两种选择，要么包含在子序列中，要么不包含在子序列中。通过减少总和，将其应用于数组中的每个元素，如果包含该元素，则搜索不包含子序列的子序列。
下面是上述方法的实现：
C++
// C++ implemenation to find all the
// subsequence whose sum is K
#include
using namespace std;
// Utility function to find the subsequences
// whose sum of the element is K
int subsetSumToK(int arr[], int n,
int output[][50], int k){
// Base Case
if (n == 0) {
if (k == 0) {
output[0][0] = 0;
return 1;
}
else {
return 0;
}
}
// Array to store the subsequences
// which includes the element arr[0]
int output1[1000][50];
// Array to store the subsequences
// which not includes the element arr[0]
int output2[1000][50];
// Recursive call to find the subsequences
// which includes the element arr[0]
int size1 = subsetSumToK(arr + 1,
n - 1, output1, k - arr[0]);
// Recursive call to find the subsequences
// which not includes the element arr[0]
int size2 = subsetSumToK(arr + 1, n - 1,
output2, k);
int i, j;
// Loop to update the results of the
// Recursive call of the function
for (i = 0; i < size1; i++) {
// Incremeing the length of
// jagged array because it includes
// the arr[0] element of the array
output[i][0] = output1[i][0] + 1;
// In the first column of the jagged
// array put the arr[0] element
output[i][1] = arr[0];
}
// Loop to update the subsequence
// in the output array
for (i = 0; i < size1; i++) {
for (j = 1; j <= output1[i][0]; j++) {
output[i][j + 1] = output1[i][j];
}
}
// Loop to update the subsequnces
// which do not include the arr[0] element
for (i = 0; i < size2; i++) {
for (j = 0; j <= output2[i][0]; j++) {
output[i + size1][j] = output2[i][j];
}
}
return size1 + size2;
}
// Function to find the subsequences
// whose sum of the element is K
void countSubsequences(int arr[], int n,
int output[][50], int k)
{
int size = subsetSumToK(arr, n, output, k);
for (int i = 0; i < size; i++) {
for (int j = 1; j <= output[i][0]; j++) {
cout << output[i][j] << " ";
}
cout << endl;
}
}
// Driver Code
int main()
{
int arr[] = {5, 12, 3, 17, 1, 18, 15, 3, 17};
int length = 9, output[1000][50], k = 6;
countSubsequences(arr, length, output, k);
return 0;
} Java
// Java implementation to find all the
// sub-sequences whose sum is K
import java.util.*;
public class SubsequenceSumK {
// Function to find the subsequences
// with given sum
public static void subSequenceSum(
ArrayList> ans,
int a[], ArrayList temp,
int k, int start)
{
// Here we have used ArrayList
// of ArrayList in in Java for
// implementation of Jagged Array
// if k < 0 then the sum of
// the current subsequence
// in temp is greater than K
if(start > a.length || k < 0)
return ;
// if(k==0) means that the sum
// of this subsequence
// is equal to K
if(k == 0)
{
ans.add(
new ArrayList(temp)
);
return ;
}
else {
for (int i = start;
i < a.length; i++) {
// Adding a new
// element into temp
temp.add(a[i]);
// After selecting an
// element from the
// array we subtract K
// by that value
subSequenceSum(ans, a,
temp, k - a[i],i+1);
// Remove the lastly
// added element
temp.remove(temp.size() - 1);
}
}
}
// Driver Code
public static void main(String args[])
{
int arr[] = {5, 12, 3, 17, 1,
18, 15, 3, 17};
int k = 6;
ArrayList> ans;
ans= new ArrayList<
ArrayList>();
subSequenceSum(ans, arr,
new ArrayList(), k, 0);
// Loop to print the subsequences
for(int i = 0; i < ans.size();
i++){
for(int j = 0;
j < ans.get(i).size(); j++){
System.out.print(
ans.get(i).get(j));
System.out.print(" ");
}
System.out.println();
}
}
} Python3
# Python3 implemenation to find all the
# subsequence whose sum is K
# Utility function to find the subsequences
# whose sum of the element is K
def subsetSumToK(arr, n, output, k):
# Base Case
if (n == 0):
if (k == 0):
output[0][0] = 0;
return 1;
else:
return 0;
# Array to store the subsequences
# which includes the element arr[0]
output1 = [[0 for j in range(50)] for i in range(1000)]
# Array to store the subsequences
# which not includes the element arr[0]
output2 = [[0 for j in range(50)] for i in range(1000)]
# Recursive call to find the subsequences
# which includes the element arr[0]
size1 = subsetSumToK(arr[1:], n - 1, output1, k - arr[0]);
# Recursive call to find the subsequences
# which not includes the element arr[0]
size2 = subsetSumToK(arr[1:], n - 1, output2, k)
# Loop to update the results of the
# Recursive call of the function
for i in range(size1):
# Incremeing the length of
# jagged array because it includes
# the arr[0] element of the array
output[i][0] = output1[i][0] + 1;
# In the first column of the jagged
# array put the arr[0] element
output[i][1] = arr[0];
# Loop to update the subsequence
# in the output array
for i in range(size1):
for j in range(1, output1[i][0]+1):
output[i][j + 1] = output1[i][j];
# Loop to update the subsequnces
# which do not include the arr[0] element
for i in range(size2):
for j in range(output2[i][0] + 1):
output[i + size1][j] = output2[i][j];
return size1 + size2;
# Function to find the subsequences
# whose sum of the element is K
def countSubsequences(arr, n, output, k):
size = subsetSumToK(arr, n, output, k);
for i in range(size):
for j in range(1, output[i][0] + 1):
print(output[i][j], end = ' ')
print()
# Driver Code
if __name__=='__main__':
arr = [5, 12, 3, 17, 1, 18, 15, 3, 17]
length = 9
output = [[0 for j in range(50)] for i in range(1000)]
k = 6;
countSubsequences(arr, length, output, k);
# This code is contributed by rutvik_56.C#
// C# implemenation to find all the
// subsequence whose sum is K
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
// Function to find the subsequences
// with given sum
public static void subSequenceSum(
List> ans, int[] a,
List temp, int k, int start)
{
// Here we have used ArrayList
// of ArrayList in in Java for
// implementation of Jagged Array
// If k < 0 then the sum of
// the current subsequence
// in temp is greater than K
if (start > a.Length || k < 0)
return;
// If (k==0) means that the sum
// of this subsequence
// is equal to K
if (k == 0)
{
ans.Add(new List(temp));
return;
}
else
{
for(int i = start;
i < a.Length; i++)
{
// Adding a new
// element into temp
temp.Add(a[i]);
// After selecting an
// element from the
// array we subtract K
// by that value
subSequenceSum(ans, a,
temp, k - a[i], i + 1);
// Remove the lastly
// added element
temp.RemoveAt(temp.Count - 1);
}
}
}
// Driver code
static public void Main ()
{
int[] arr = { 5, 12, 3, 17, 1,
18, 15, 3, 17 };
int k = 6;
List> ans = new List>();
subSequenceSum(ans, arr,
new List(), k, 0);
// Loop to print the subsequences
for(int i = 0; i < ans.Count; i++)
{
for(int j = 0; j < ans[i].Count; j++)
{
Console.Write(ans[i][j] + " ");
}
Console.WriteLine();
}
}
}
// This code is contributed by offbeat 输出：
5 1
3 3