連續揹包問題

鑑於物品為 (value, weight)，我們需要將它們放置在容量為 k 的揹包（容器）中。注意！ 我們可以打破物品以最大化價值！

示例輸入：

values[] = [1, 4, 5, 2, 10]
weights[] = [3, 2, 1, 2, 4]
k = 8

預期輸出：

maximumValueOfItemsInK = 20;

演算法：

1) Sort values and weights by value/weight.
   values[] = [5, 10, 4, 2, 1]
   weights[] = [1, 4, 2, 2, 3]
2) currentWeight = 0; currentValue = 0;
3) FOR i = 0; currentWeight < k && i < values.length; i++ DO:
       IF k - currentWeight < weights[i] DO
           currentValue = currentValue + values[i];
           currentWeight = currentWeight + weights[i];
       ELSE
           currentValue = currentValue + values[i]*(k - currentWeight)/weights[i]
           currentWeight = currentWeight + weights[i]*(k - currentWeight)/weights[i]
       END_IF
   END_FOR
   PRINT "maximumValueOfItemsInK = " + currentValue;