我试图写一个C code生成所有可能的分区(到2个或更多份)的不同的的给定数目的元素。一个给定的分区的所有数字的总和应等于给定数。例如,对于输入每组6 ,系统具有2个或更多个元件以不同元件的所有可能的分区是:
I am trying to write a C code to generate all possible partitions (into 2 or more parts) with distinct elements of a given number. The sum of all the numbers of a given partition should be equal to the given number. For example, for input n = 6, all possible partitions having 2 or more elements with distinct elements are:
- 1,5
- 在1,2,3
- 2,4
我觉得递归的方法应该可行,但我不能把不同元素的添加约束的照顾。伪code或样品code在C / C ++ / Java的将是极大的AP preciated。
I think a recursive approach should work, but I am unable to take care of the added constraint of distinct elements. A pseudo code or a sample code in C/C++/Java would be greatly appreciated.
谢谢!
编辑::如果它使事情变得更容易,我可以忽略其至少有2个单元分区的限制。这将允许号码自身被添加到列表中(例如,6本身将是微不足道的,但有效的分区)。
If it makes things easier, I can ignore the restriction of the partitions having atleast 2 elements. This will allow the number itself to be added to the list (eg, 6 itself will be a trivial but valid partition).
推荐答案我勾勒这个解决方案(也可以美化和优化),这不应该产生重复的:
I sketched this solution (it can be beautified and optimized) that shouldn't generate duplicates:
void partitions(int target, int curr, int* array, int idx) { if (curr + array[idx] == target) { for (int i=0; i <= idx; i++) cout << array[i] << " "; cout << endl; return; } else if (curr + array[idx] > target) { return; } else { for(int i = array[idx]+1; i < target; i++) { array[idx+1] = i; partitions(target, curr + array[idx], array, idx+1); } } } int main(){ int array[100]; int N = 6; for(int i = 1; i < N; i++) { array[0] = i; partitions(N, 0, array, 0); } }更多推荐
生成一个号码的所有不同的分区
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