Wolfram mathematica Mathematica中的下料变量

所以我对Mathematica还很陌生，我正在努力学习用函数的方式解决问题。我要解决的问题是列出我可以对列表中的元素求和的方法（有重复），所以求和是leq到某个值。下面的代码很好地解决了这个问题

i = {7.25, 7.75, 15, 19, 22};
m = 22;
getSum[l_List, n_List] := Total[Thread[{l, n}] /. {x_, y_} -> x y];
t = Prepend[Map[Range[0, Floor[m/#]] &, i], List];
Outer @@ %;
Flatten[%, ArrayDepth[%] - 2];
Map[{#, getSum[i, #]} &, %];
DeleteCases[%, {_, x_} /; x > m || x == 0];
TableForm[Flatten /@ SortBy[%, Last], 0, 
 TableHeadings -> {None, Append[i, "Total"]}]

---

然而，代码检查了很多不必要的情况，如果列表中m越高，则可能会出现问题。我的问题很简单，在效率和代码优雅方面，解决这个问题最像数学的方法是什么

一个简单但不是最佳的方法是：

sol = Reduce[Dot[i, {a, b, c, d, e}] <= m, {a, b, c, d, e}, Integers];

怎么样

recurr[numbers_, boundary_] :=

  Reap[memoryRecurr[0, {}, numbers, boundary]][[2, 1]];

memoryRecurr[_, _, {}, _] := Null;

memoryRecurr[sum_, numbers_, restNumbers_, diff_] :=
  (
   Block[
     {presentNumber = First[restNumbers], restRest = Rest[restNumbers]}
     ,
     If[
      presentNumber <= diff
      ,
      Block[{
        newNumbers = Append[numbers, presentNumber],
        newSum = sum + presentNumber
        },
       Sow[{newNumbers, newSum}];

       memoryRecurr[
        newSum,
        newNumbers,
        restRest,
        diff - presentNumber
        ];
       ]
      ];
     memoryRecurr[sum, numbers, restRest, diff]
     ];

   );

->

---

我想投赞成票，但你的总代表现在看起来比加10好得多。@HighPerformanceMark我必须同意，我会接受精神上的支持。代码确实非常简洁！然而，与我的解决方案相比，它似乎非常慢，这正是我想要改进的！啊，其他人破坏了你代表的优雅，所以我现在投票。在这里和Mathematica.StackExchange上交叉张贴没有多大意义-两个地方的大多数人都是相同的，只有我们中的大多数人切换到了那个。

recurr[numbers_, boundary_] :=

  Reap[memoryRecurr[0, {}, numbers, boundary]][[2, 1]];

memoryRecurr[_, _, {}, _] := Null;

memoryRecurr[sum_, numbers_, restNumbers_, diff_] :=
  (
   Block[
     {presentNumber = First[restNumbers], restRest = Rest[restNumbers]}
     ,
     If[
      presentNumber <= diff
      ,
      Block[{
        newNumbers = Append[numbers, presentNumber],
        newSum = sum + presentNumber
        },
       Sow[{newNumbers, newSum}];

       memoryRecurr[
        newSum,
        newNumbers,
        restRest,
        diff - presentNumber
        ];
       ]
      ];
     memoryRecurr[sum, numbers, restRest, diff]
     ];

   );

recurr[{1, 2, 3, 4, 5}, 7]

{{{1}, 1}, {{1, 2}, 3}, {{1, 2, 3}, 6}, {{1, 2, 4}, 7}, {{1, 3}, 
  4}, {{1, 4}, 5}, {{1, 5}, 6}, {{2}, 2}, {{2, 3}, 5}, {{2, 4}, 
  6}, {{2, 5}, 7}, {{3}, 3}, {{3, 4}, 7}, {{4}, 4}, {{5}, 5}}