你如何从给定的N个数字中测试所有可能的加法组合，使它们加起来得到给定的最终数字?
一个简单的例子:
要添加的数字集:N ={1,5,22,15,0，…} 期望结果:12345
当前回答
我想我应该用这个问题的答案，但我不能，所以这是我的答案。它使用的是《计算机程序的结构和解释》中答案的修改版本。我认为这是一个更好的递归解，应该更能取悦纯粹主义者。
我的答案是用Scala(如果我的Scala很烂，我很抱歉，我刚刚开始学习)。findsumcombination的疯狂之处在于对递归的原始列表进行排序和惟一，以防止欺骗。
def findSumCombinations(target: Int, numbers: List[Int]): Int = {
cc(target, numbers.distinct.sortWith(_ < _), List())
}
def cc(target: Int, numbers: List[Int], solution: List[Int]): Int = {
if (target == 0) {println(solution); 1 }
else if (target < 0 || numbers.length == 0) 0
else
cc(target, numbers.tail, solution)
+ cc(target - numbers.head, numbers, numbers.head :: solution)
}
使用它:
> findSumCombinations(12345, List(1,5,22,15,0,..))
* Prints a whole heap of lists that will sum to the target *
其他回答
我想我应该用这个问题的答案，但我不能，所以这是我的答案。它使用的是《计算机程序的结构和解释》中答案的修改版本。我认为这是一个更好的递归解，应该更能取悦纯粹主义者。
我的答案是用Scala(如果我的Scala很烂，我很抱歉，我刚刚开始学习)。findsumcombination的疯狂之处在于对递归的原始列表进行排序和惟一，以防止欺骗。
def findSumCombinations(target: Int, numbers: List[Int]): Int = {
cc(target, numbers.distinct.sortWith(_ < _), List())
}
def cc(target: Int, numbers: List[Int], solution: List[Int]): Int = {
if (target == 0) {println(solution); 1 }
else if (target < 0 || numbers.length == 0) 0
else
cc(target, numbers.tail, solution)
+ cc(target - numbers.head, numbers, numbers.head :: solution)
}
使用它:
> findSumCombinations(12345, List(1,5,22,15,0,..))
* Prints a whole heap of lists that will sum to the target *
这个问题的解决方案在互联网上已经出现过无数次了。这个问题叫做硬币兑换问题。你可以在http://rosettacode.org/wiki/Count_the_coins上找到答案，在http://jaqm.ro/issues/volume-5,issue-2/pdfs/patterson_harmel.pdf上找到数学模型(或谷歌硬币变化问题)。
顺便说一下，Tsagadai的Scala解决方案很有趣。本例生成1或0。作为一个副作用，它在控制台上列出了所有可能的解决方案。它显示解决方案，但无法以任何方式使其可用。
为了尽可能有用，代码应该返回一个List[List[Int]]，以允许获得解决方案的数量(列表列表的长度)，“最佳”解决方案(最短的列表)，或所有可能的解决方案。
这里有一个例子。它效率很低，但很容易理解。
object Sum extends App {
def sumCombinations(total: Int, numbers: List[Int]): List[List[Int]] = {
def add(x: (Int, List[List[Int]]), y: (Int, List[List[Int]])): (Int, List[List[Int]]) = {
(x._1 + y._1, x._2 ::: y._2)
}
def sumCombinations(resultAcc: List[List[Int]], sumAcc: List[Int], total: Int, numbers: List[Int]): (Int, List[List[Int]]) = {
if (numbers.isEmpty || total < 0) {
(0, resultAcc)
} else if (total == 0) {
(1, sumAcc :: resultAcc)
} else {
add(sumCombinations(resultAcc, sumAcc, total, numbers.tail), sumCombinations(resultAcc, numbers.head :: sumAcc, total - numbers.head, numbers))
}
}
sumCombinations(Nil, Nil, total, numbers.sortWith(_ > _))._2
}
println(sumCombinations(15, List(1, 2, 5, 10)) mkString "\n")
}
运行时，它显示:
List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)
List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2)
List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2)
List(1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2)
List(1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2)
List(1, 1, 1, 1, 1, 2, 2, 2, 2, 2)
List(1, 1, 1, 2, 2, 2, 2, 2, 2)
List(1, 2, 2, 2, 2, 2, 2, 2)
List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5)
List(1, 1, 1, 1, 1, 1, 1, 1, 2, 5)
List(1, 1, 1, 1, 1, 1, 2, 2, 5)
List(1, 1, 1, 1, 2, 2, 2, 5)
List(1, 1, 2, 2, 2, 2, 5)
List(2, 2, 2, 2, 2, 5)
List(1, 1, 1, 1, 1, 5, 5)
List(1, 1, 1, 2, 5, 5)
List(1, 2, 2, 5, 5)
List(5, 5, 5)
List(1, 1, 1, 1, 1, 10)
List(1, 1, 1, 2, 10)
List(1, 2, 2, 10)
List(5, 10)
sumcombination()函数可以单独使用，并且可以进一步分析结果以显示“最佳”解决方案(最短的列表)或解决方案的数量(列表的数量)。
请注意，即使这样，需求也可能无法完全满足。解决方案中每个列表的顺序可能是重要的。在这种情况下，每个列表都必须重复它的元素组合的次数。或者我们只对不同的组合感兴趣。
例如，我们可以考虑List(5,10)应该给出两种组合:List(5,10)和List(10,5)。对于List(5,5,5)，它可以给出三种组合，也可以只给出一种组合，这取决于需求。对于整数，这三种排列是等价的，但如果我们处理的是硬币，就像在“硬币更换问题”中一样，它们就不一样了。
Also not stated in the requirements is the question of whether each number (or coin) may be used only once or many times. We could (and we should!) generalize the problem to a list of lists of occurrences of each number. This translates in real life into "what are the possible ways to make an certain amount of money with a set of coins (and not a set of coin values)". The original problem is just a particular case of this one, where we have as many occurrences of each coin as needed to make the total amount with each single coin value.
Excel VBA版本如下。我需要在VBA中实现这一点(不是我的偏好，不要评判我!)，并使用本页上的答案作为方法。我上传以防其他人也需要VBA版本。
Option Explicit
Public Sub SumTarget()
Dim numbers(0 To 6) As Long
Dim target As Long
target = 15
numbers(0) = 3: numbers(1) = 9: numbers(2) = 8: numbers(3) = 4: numbers(4) = 5
numbers(5) = 7: numbers(6) = 10
Call SumUpTarget(numbers, target)
End Sub
Public Sub SumUpTarget(numbers() As Long, target As Long)
Dim part() As Long
Call SumUpRecursive(numbers, target, part)
End Sub
Private Sub SumUpRecursive(numbers() As Long, target As Long, part() As Long)
Dim s As Long, i As Long, j As Long, num As Long
Dim remaining() As Long, partRec() As Long
s = SumArray(part)
If s = target Then Debug.Print "SUM ( " & ArrayToString(part) & " ) = " & target
If s >= target Then Exit Sub
If (Not Not numbers) <> 0 Then
For i = 0 To UBound(numbers)
Erase remaining()
num = numbers(i)
For j = i + 1 To UBound(numbers)
AddToArray remaining, numbers(j)
Next j
Erase partRec()
CopyArray partRec, part
AddToArray partRec, num
SumUpRecursive remaining, target, partRec
Next i
End If
End Sub
Private Function ArrayToString(x() As Long) As String
Dim n As Long, result As String
result = "{" & x(n)
For n = LBound(x) + 1 To UBound(x)
result = result & "," & x(n)
Next n
result = result & "}"
ArrayToString = result
End Function
Private Function SumArray(x() As Long) As Long
Dim n As Long
SumArray = 0
If (Not Not x) <> 0 Then
For n = LBound(x) To UBound(x)
SumArray = SumArray + x(n)
Next n
End If
End Function
Private Sub AddToArray(arr() As Long, x As Long)
If (Not Not arr) <> 0 Then
ReDim Preserve arr(0 To UBound(arr) + 1)
Else
ReDim Preserve arr(0 To 0)
End If
arr(UBound(arr)) = x
End Sub
Private Sub CopyArray(destination() As Long, source() As Long)
Dim n As Long
If (Not Not source) <> 0 Then
For n = 0 To UBound(source)
AddToArray destination, source(n)
Next n
End If
End Sub
输出(写入立即窗口)应该是:
SUM ( {3,8,4} ) = 15
SUM ( {3,5,7} ) = 15
SUM ( {8,7} ) = 15
SUM ( {5,10} ) = 15
c#版本的@msalvadores代码的答案
void Main()
{
int[] numbers = {3,9,8,4,5,7,10};
int target = 15;
sum_up(new List<int>(numbers.ToList()),target);
}
static void sum_up_recursive(List<int> numbers, int target, List<int> part)
{
int s = 0;
foreach (int x in part)
{
s += x;
}
if (s == target)
{
Console.WriteLine("sum(" + string.Join(",", part.Select(n => n.ToString()).ToArray()) + ")=" + target);
}
if (s >= target)
{
return;
}
for (int i = 0;i < numbers.Count;i++)
{
var remaining = new List<int>();
int n = numbers[i];
for (int j = i + 1; j < numbers.Count;j++)
{
remaining.Add(numbers[j]);
}
var part_rec = new List<int>(part);
part_rec.Add(n);
sum_up_recursive(remaining,target,part_rec);
}
}
static void sum_up(List<int> numbers, int target)
{
sum_up_recursive(numbers,target,new List<int>());
}
非常有效的算法，使用我几年前用c++写的表格。
如果你设置PRINT 1，它将打印所有的组合(但它不会使用有效的方法)。
它非常高效，在不到10毫秒的时间内计算了超过10^14个组合。
#include <stdio.h>
#include <stdlib.h>
//#include "CTime.h"
#define SUM 300
#define MAXNUMsSIZE 30
#define PRINT 0
long long CountAddToSum(int,int[],int,const int[],int);
void printr(const int[], int);
long long table1[SUM][MAXNUMsSIZE];
int main()
{
int Nums[]={3,4,5,6,7,9,13,11,12,13,22,35,17,14,18,23,33,54};
int sum=SUM;
int size=sizeof(Nums)/sizeof(int);
int i,j,a[]={0};
long long N=0;
//CTime timer1;
for(i=0;i<SUM;++i)
for(j=0;j<MAXNUMsSIZE;++j)
table1[i][j]=-1;
N = CountAddToSum(sum,Nums,size,a,0); //algorithm
//timer1.Get_Passd();
//printf("\nN=%lld time=%.1f ms\n", N,timer1.Get_Passd());
printf("\nN=%lld \n", N);
getchar();
return 1;
}
long long CountAddToSum(int s, int arr[],int arrsize, const int r[],int rsize)
{
static int totalmem=0, maxmem=0;
int i,*rnew;
long long result1=0,result2=0;
if(s<0) return 0;
if (table1[s][arrsize]>0 && PRINT==0) return table1[s][arrsize];
if(s==0)
{
if(PRINT) printr(r, rsize);
return 1;
}
if(arrsize==0) return 0;
//else
rnew=(int*)malloc((rsize+1)*sizeof(int));
for(i=0;i<rsize;++i) rnew[i]=r[i];
rnew[rsize]=arr[arrsize-1];
result1 = CountAddToSum(s,arr,arrsize-1,rnew,rsize);
result2 = CountAddToSum(s-arr[arrsize-1],arr,arrsize,rnew,rsize+1);
table1[s][arrsize]=result1+result2;
free(rnew);
return result1+result2;
}
void printr(const int r[], int rsize)
{
int lastr=r[0],count=0,i;
for(i=0; i<rsize;++i)
{
if(r[i]==lastr)
count++;
else
{
printf(" %d*%d ",count,lastr);
lastr=r[i];
count=1;
}
}
if(r[i-1]==lastr) printf(" %d*%d ",count,lastr);
printf("\n");
}
