计算矩阵中的交叉对角元素的总和 [英] Calculate the sum of cross-diagonal elements in a matrix

问题描述

我想把矩阵中的交叉对角元素加在一起。例如，我有一个3 * 3的矩阵是二维的，我想将它转换为一维：

   - ----------------- 
 | 1 | 2 | 3 | 
 ------------------- 
 A = | 4 | 5 | 6 | 
 ------------------- 
 | 7 | 8 | 9 | 
 ------------------- 
  

<

pre code $ ____ ____ ____ ____
B = | 1 | 6 | 15 | 14 | 9 |
| ____ | ____ | ____ | ____ | ____ |

第一个交叉对角线 A [0] [0] 将被复制到 B [0] 。

然后下一个交叉对角元素 A [1] [0] 和 A [0] [1] 将被添加并复制到 B [1] ， ie 4和2将被添加。

然后下一个交叉对角元素 code> A [2] [0] 和 A [1] [1] 和 A [0 ] [2] 将被添加并复制到 B [2] ，即 7,5和3将会添加。

等等...

解决方案

对于每个对角线，行索引和列索引的总和等于B数组的索引。基于这个事实，你可以像这样做一个算法：

$ pre code> //假设矩阵的宽度和长度是N
//很好，你对这个想法的范围有一些想法，试试看你自己？
//对于（int i = 0; i （int j = 0; j <= i），肯定它应该是N
的函数; j ++）{//考虑为什么j应该在范围内（0，i）？
//一些累计在这里添加
}
}

I want to add together the cross-diagonal elements in a matrix. For example, I have a 3*3 Matrix which is two dimensional, I want to convert it to one dimensional:

      -------------------
      |  1  |  2  |  3  |
      -------------------
 A=   |  4  |  5  |  6  |
      -------------------
      |  7  |  8  |  9  |
      -------------------

final output will be,

     ____ ____ ____ ____ ____
 B= |1   | 6  | 15 | 14 |  9 |
    |____|____|____|____|____|

First cross-diagonal A[0][0] will be copied to B[0].

Then the next cross-diagonal elements A[1][0] and A[0][1] will be added and copied to B[1], i.e. 4 and 2 will be added.

Then the next cross-diagonal elements A[2][0] and A[1][1] and A[0][2] will be added and copied to B[2], i.e. 7, 5 and 3 will be added.

And so on...

解决方案

Notice that for each diagonal, the sum of row-index and column-index is equal to the index of B array. Based on this fact, you can make a algorithm like this:

// assuming the width and length of the Matrix is N
// it's good you have some ideas of the range of idea, try figure it out by yourself? 
// definitely it should be a function of N
for (int i=0;i<F(N);i++) { 
  for (int j=0;j<=i;j++) { // consider why j should be in range (0,i) ?
    // some cumulatively add here
  }
} 

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