[英]Finding subset sum and returning it and the subset
我對 Java 和編碼相對較新,目前正在嘗試解決子集問題。
目標是讓用戶輸入數字的數量,然后輸入每個數字,直到指定的數量,計算機相應地要求每個第 n 個數字。
之后計算機應計算所有子集的總和,並返回最接近 pi 的子集,以及同一行中這種形式 [x,y,z] 的子集。
我很好地管理了第一部分,盡管為了方便起見,它可能已經通過開關盒進行了改進。 它將輸入數字添加到數組中
但我在這個問題的第二部分苦苦掙扎,我不知道如何推進/安排代碼,以便輸出所需的結果。 我得到的建議是,一個包含 n 個元素的集合的 for 循環:
for a from 0 to 2^n
for i from 0 to n
when the binary representation of x on has a 1 at the i-th position
add data[i] to solution.
據說這應該找到數組的所有子集。 之后我應該添加每個元素,檢查與 pi 的距離是否減小並將元素添加到解決方案集中。 或者至少這是目標,但我的代碼不起作用,因為我不知道從哪里開始安排它。 我也不知道用什么來初始化 bestsum,或者二進制表示算法是如何工作的,或者如何按順序將元素添加到解決方案數組中。
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("How many numbers should be read? ");
int count = input.nextInt();
double data[] = new double[count];
double solution[] = new double[count];
double bestsum = Double.POSITIVE_INFINITY;
for (int i = 0; i < count; i++) {
if (i % 10 == 1) {
System.out.println("Enter " + i + "st number: ");
data[i] = input.nextDouble();
} else if (i % 10 == 2) {
System.out.println("Enter " + i + "nd number: ");
data[i] = input.nextDouble();
} else if (i % 10 == 3) {
System.out.println("Enter " + i + "rd number: ");
data[i] = input.nextDouble();
} else {
System.out.println("Enter " + i + "th number: ");
data[i] = input.nextDouble();
}
}
for (long x = 0; x <= Math.pow(2,20); x++) {
for (int y = 0; y <= 20; y++) {
if (((x >> y) & 1) == 1) {
data[y] = solution[y];
}
if (abs(bestsum + solution[y] - PI) < bestsum)
bestsum = bestsum + solution[y];
}
}
System.out.println("Best sum: ");
String sol = solution.toString();
System.out.println(bestsum + sol);
這是我對這個問題的解決方案。
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("How many numbers should be read? ");
int count = input.nextInt();
Double data[] = new Double[count];
Double solution[] = new Double[count];
Double bestsum = 0.0; //sum for an emptyset
Set<Double> bestSet= Collections.emptySet();
Set<Set<Double>> powerSet = new LinkedHashSet<>();
for (int i = 0; i < count; i++) {
if (i == 0) {
System.out.println("Enter " + (i+1) + "st number: ");
data[i] = input.nextDouble();
} else if (i == 1) {
System.out.println("Enter " + (i+1) + "nd number: ");
data[i] = input.nextDouble();
} else if (i == 2) {
System.out.println("Enter " + (i+1) + "rd number: ");
data[i] = input.nextDouble();
} else {
System.out.println("Enter " + (i+1) + "th number: ");
data[i] = input.nextDouble();
}
}
for (int i = 0; i < (1<<count); i++)
{
int m = 1; // m is used to check set bit in binary representation.
Set<Double> currentSet = new LinkedHashSet<>();
for (int j = 0; j < count; j++)
{
if ((i & m) > 0)
{
currentSet.add(data[j]);
}
m = m << 1;
}
powerSet.add(currentSet);
}
Iterator<Set<Double>> iterator = powerSet.iterator();
iterator.next(); //lets skip the first set which is an emptySet
for (int i = 1; i < (1<<count); i++) {
Double sum=0.0;
Set<Double> currentSet = iterator.next();
sum = currentSet.stream().collect(Collectors.summingDouble(Double::doubleValue));
if(Math.abs(sum-Math.PI)<Math.abs(bestsum-Math.PI)) {
bestSet = currentSet;
bestsum = sum;
}
}
System.out.println("powerSet: " + powerSet);
System.out.println("solution: " + bestSet);
System.out.println("bestsum: " + bestsum);
}
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