总和在一个数组，C的连续元素++ [英] Sum of consecutive elements in an array, C++

问题描述

假设我有包括n个元素的数组。

Suppose that I have an array consisting of n elements.

1 2 3 4 5 6 ... n

我需要找到一个方法来提取使用C ++这个数组的连续元素的总和。 像这样的：

I need to find a way to extract the sums of consecutive elements in this array using C++. Like this:

1, 2, 3,...n, 1+2, 2+3, 3+4,...(n-1)+n, 1+2+3, 2+3+4,...(n-2)+(n-1)+n,...1+2+3...n

到目前为止，我想通了，我需要通过总结一定数量的在每次运行时的元素在这个数组中迭代。我不知道是否有可能实现的算法，我上面的解释。有可能是一个更好的解决方案，但是这是最好的我能想出。

So far I figured out I need to iterate through this array by summing certain number of elements on each run. I am not sure if it is possible to implement the algorithm I explained above. There might be a better solution but this is the best I could come up with.

推荐答案

让我们检查的情况下有4个元素：

Let's inspect case with 4 elements:

{1,3,4,5, // from original array
 4,7,9, // sum of 2 consecutive elements
 8,12, // sum of 3
 13} // sum of 4

正如你所看到的每一个部分对于N之和数组的大小较低由原来的数组（N-1）。因此，你需要目标大小的数组：N +（N-1）+（N-2）+ ... 1 - 这是N *（1 + N）/ 2

As you can see every part for N sum array is of size lower from original array by (N-1). So you need target array of size: N + (N-1) + (N-2) + ... 1 - which is N*(1+N)/2

int* createSumArray(int* arr, int size)
{
   int ti = 0; // target index
   int* ta = new int[size*(size+1)/2];
   for (int s = 1; s <= size; ++s) // how many elements to sum
   {
      for (int si = 0; si < size + 1 - s; ++si)
      {
          ta[ti] = 0;
          for (int i = si; i < si + s; ++i)
            ta[ti] += arr[i];
          ++ti;
      } 
   }
   return ta;
}

请参阅测试在 ideone

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