我目前正在备考，我正在尝试处理动态矩阵。我遇到了一个关于计算矩阵的每条对角线之和的问题，该矩阵的值和大小由用户选择。 我的程序的目的是打印，多亏了一个函数，其参数是矩阵及其大小，每个对角线和的值。我将向您展示代码并对其进行深入描述。
----------------
| 52 | 35 | 5 | Example of matrix.
---------------- Imagine the first diagonal to be the one which goes right-to-left
| 2 | 71 | 1 | and only consists in the number "47".
---------------- The second diagonal would be the one which goes right-to-left and
| 3 | 60 | 25 | consists in the number "15" and "79".
---------------- So to get the sum of the second diagonal it would be:
| 79 | 55 | 98 |
---------------- sum = m[n_rows - 1][diag - 2] + m[n_rows - 2][diag - 1]
| 47 | 15 | 66 |
---------------- When diag > columns, in order to avoid error regarding matrix size,
I should lower the quantity "n_rows - 1" by the quantity "diag - n_columns".
根据我的描述，这是我想做的:
void diag_matrix(int** m, int righe, int colonne){//righe = rows, colonne = columns.
//M is the matrix.
// diag is the number of the diagonal I'm considering.
for(int diag = 1; diag < (righe + colonne); diag++){
int sum = 0;// the sum
int i = 0;// the counter of the cicle
int l = 0;// this is the value to riallign the row in case diag > column
int temp = diag;//I use this variable not to modify the value of diag.
// What I want is: when the column-index/row-index of the matrix reaches 0, the cicle will interrupt (after final iteration);
while(righe - i - l - 1 > 0 || diag - 1 - i > 0){
if (diag > colonne){//this condition changes l-value only if diag value is greater than column. Explanation outside the code
l = diag - colonne;//this is the value to subtract to row-index
temp = colonne;//this position is necessary to set column-index to its maxium.
}
sum = sum + m[righe - 1 - l - i][temp -1 - i];//pretty clear I think.
i++;//the i is incremented by one.
}// end of while-statement
cout << "Somma Diagonale " << diag << " = " << sum << ".\n";
}// end of for-statement
}//end of function declaration
显然是行不通的，但我想不出问题所在。
最佳答案
(这里本来有一段话，但仔细一看，你并没有犯它说的错误。)
由于您没有发布到代码审查，这里有一个解决方案而不是详细的代码审查。 (如果你想让原来的方法起作用，我建议在调试器中单步执行它并检查你的变量首先在哪里获得错误值。)它有很多样板来编译和运行，但是您最感兴趣的部分是 diag_sums() 及其注释。
这里的一个想法是使用 OOP 来自动检查数组访问的边界。后者对于捕获逐一错误等非常重要。如果你愿意，你可以在生产中关闭它，但你真的不想在你的程序有缓冲区溢出时关闭警告。这里的其他优化包括数据访问的局部性，以及操作的强度降低:我们可以简单地预先计算每条对角线的长度，而不是检查每次迭代是否碰到右边缘和下边缘。
由于M行矩阵a的对角线数k的定义等价于:所有元素a[i] [j] 使得 M - k = i - j，该算法确保通过保持不变性来保持正确性，每当我们将 i 和 j 都加 1 时，从 i 或 j为0，每当i = M或j = N时停止，即遍历每一步从左边缘或上边缘到右边缘或下边缘的对角线，以先到者为准。
#include <assert.h>
#include <iostream>
#include <stddef.h>
#include <stdlib.h>
#include <utility>
#include <vector>
using std::cin;
using std::cout;
template <typename T>
class matrix {
public:
matrix( const ptrdiff_t rows,
const ptrdiff_t cols,
std::vector<T>&& elems )
: rows_(rows), cols_(cols), elems_(elems)
{
assert( rows_ > 0 );
assert( cols_ > 0 );
assert( elems_.size() == static_cast<size_t>(rows_*cols_) );
}
matrix( const ptrdiff_t rows,
const ptrdiff_t cols,
const std::vector<T>& elems )
: matrix( rows, cols, std::move(std::vector<T>(elems)) )
{}
matrix( const matrix<T>& ) = default;
matrix( matrix<T>&& ) = default;
matrix& operator= ( const matrix<T>& ) = default;
matrix& operator= ( matrix<T>&& ) = default;
T& operator() ( const ptrdiff_t m, const ptrdiff_t n )
{
assert( m >= 0 && m < rows_ );
assert( n >= 0 && n < cols_ );
return elems_[static_cast<size_t>(m*cols_ + n)];
}
const T& operator() ( const ptrdiff_t m, const ptrdiff_t n ) const
{
/* Because this call does not modify any data, and the only reason the
* member function above cannot be const is that it returns a non-const
* reference to an element of elems, casting away the const qualifier
* internally and then returning a const reference is a safe way to
* re-use the code.
*/
matrix<T>& nonconst = *const_cast<matrix<T>*>(this);
return nonconst(m,n);
}
ptrdiff_t rows() const { return rows_; }
ptrdiff_t cols() const { return cols_; }
private:
ptrdiff_t rows_;
ptrdiff_t cols_;
std::vector<T> elems_;
};
template<typename T>
std::ostream& operator<< ( std::ostream& out, const matrix<T>& x )
/* Boilerplate to print a matrix. */
{
const ptrdiff_t m = x.rows(), n = x.cols();
for ( ptrdiff_t i = 0; i < m; ++i ) {
out << x(i,0);
for ( ptrdiff_t j = 1; j < n; ++j )
out << ' ' << x(i,j);
out << '\n';
} // end for
return out;
}
using elem_t = int;
std::vector<elem_t> diag_sums( const matrix<elem_t>& a )
/* Return a vector of all the diagonal sums of a.
*
* The first diagonal sum is a(rows-1,0)
* The second is a(rows-2,0) + a(rows-1,1)
* The third is a(rows-3,0) + a(rows-2,1) + a(rows-1,2)
* And so on. I.e., the kth diagonal is the sum of all elements a(i,j) such
* that i - j == rows - k.
*
* If a is a M×N matrix, there are M diagonals starting in column zero, and
* N-1 diagonals (excluding the one containing a(0,0) so we don't count it
* twice) starting in row 0. We process them bottom to top, then left to
* right.
*
* The number of elements in a diagonal starting at a(i,0) is min{M-i, N}. The
* number of elements in a diagonal starting at a(0,j) is min{M, N-j}. This is
* because a diagonal stops at either the bottom edge or the left edge of a.
*/
{
const ptrdiff_t m = a.rows(), n = a.cols();
std::vector<elem_t> result;
result.reserve( static_cast<size_t>(m + n - 1) );
for ( ptrdiff_t i = m-1; i > 0; --i ) {
elem_t sum = 0;
const ptrdiff_t nk = (m-i) < n ? (m-i) : n;
for ( ptrdiff_t k = 0; k < nk; ++k )
sum += a(i+k, k);
result.emplace_back(sum);
} // end for i
for ( ptrdiff_t j = 0; j < n; ++j ) {
elem_t sum = 0;
const ptrdiff_t nk = m < (n-j) ? m : (n-j);
for ( ptrdiff_t k = 0; k < nk; ++k )
sum += a(k, j+k);
result.emplace_back(sum);
} // end for j
return result;
}
matrix<elem_t> read_input_matrix( const int row, const int column )
/* Reads in row*column consecutive elements from cin and packs them into a
* matrix<elem_t>.
*/
{
assert(row > 0);
assert(column > 0);
const ptrdiff_t nelements = row*column;
assert(nelements > 0); // Check for overflow.
std::vector<elem_t> result;
result.reserve(static_cast<size_t>(nelements));
for ( ptrdiff_t i = nelements; i > 0; --i ) {
int x;
cin >> x;
assert(cin.good());
result.push_back(x);
}
return matrix<elem_t>( row,
column,
std::move(result) );
}
template<typename T>
bool print_sequence( const T& container )
/* Prints the contents of a container in the format
* "{47, 94, 124, 160, 148, 36, 5}".
*/
{
cout << "{";
if ( container.begin() != container.end() )
cout << *container.begin();
for ( auto it = container.begin() + 1; it < container.end(); ++it )
cout << ", " << *it;
cout << "}\n";
return cout.good();
}
/* A simple test driver that reads in the number of rows, the number of
* columns, and then row*columns int values, from standard input. It
* then passes the result to diag_matrix(), E.g.:
*
* 5 3
* 52 35 5
* 2 71 1
* 3 60 25
* 79 55 98
* 47 15 66
*/
int main()
{
int rows, columns;
cin >> rows;
cin >> columns;
assert(cin.good());
const matrix<elem_t> input_matrix = read_input_matrix( rows, columns );
// cout << input_matrix; // Instrumentation.
const std::vector<elem_t> sums = diag_sums(input_matrix);
print_sequence(sums);
return EXIT_SUCCESS;
}
您也可以只执行 print_sequence(diag_sums(read_input_matrix( rows, columns )))。
关于c++ - 矩阵中独立对角线的总和，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/48570937/