双数据类型的子集总和？

Question

我有以下适用于整数的子集合的代码。如何将此代码扩展为双数据类型输入？例如，当输入是1.01,2.65,3.08,4.07,5.12（比如说）和输出是15.62（比如说）时，如何扩展这个相同的代码。这些输入和输出都是示例，即使它们改变代码也应该工作。

// A Java program to count all subsets with given sum. 
import java.util.ArrayList; 
public class subset_sum 
{ 
// 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<Integer> 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<Integer> 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<Integer> 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<n; ++i) 
    { 
     dp[i][0] = true; 
    } 

    // Sum arr[0] can be achieved with single element 
    if (arr[0] <= sum) 
     dp[0][arr[0]] = true; 

    // Fill rest of the entries in dp[][] 
    for (int i = 1; i < n; ++i) 
     for (int j = 0; j < sum + 1; ++j) 
      dp[i][j] = (arr[i] <= j) ? (dp[i-1][j] || 
        dp[i-1][j-arr[i]]) 
        : dp[i - 1][j]; 
    if (dp[n-1][sum] == false) 
    { 
     System.out.println("There are no subsets with" + 
       " sum "+ sum); 
     return; 
    } 

    // Now recursively traverse dp[][] to find all 
    // paths from dp[n-1][sum] 
    ArrayList<Integer> 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); 
} 
} 

输出：[4，3，2，1] [5,3，2] [5,4，1]

Answer 1

我通过简单地通过计算转换双到整数发现这个问题的答案小数位数乘以100（说）作为算法使用此外，这种变化不会影响最终值在这种情况下，我将最终值除以100以获得结果并以双数据类型显示它