Question

我尝试编写此代码以根据奇数的用户输入输出奇数幻方。当我输入1或3时，它工作正常。每当我输入任何高于5,7,9,11等的东西时，程序会在我按下回车时崩溃。我已经查看了我的代码，但我无法确定问题所在。我没有收到错误消息。

小注意：如果你知道什么是魔方，那么我的算法（由英语教授给我们翻译成C ++）不会输出正确的值，因为它们并非都加起来相同的数字

#include <iostream>
#include <iomanip>

using namespace std;


int main()
{
int n; //n = order
cout << "Enter an odd integer for the order of the Magic Square: ";
cin >> n;
cout << endl;
if(n%2 == 0)    //only allows program to accept odd numbers
    {
    cout << "The number you have entered is not odd" << endl;
    return 0;
    }
int x, y; //x and y access the columns and rows of the following matrix
int magicsquare[n][n];  //creates a n by n matrix to set up magic square
int counter, square = n*n; //square is upper boundary

for(x=0; x<n; x++)      //initialize all spaces in matrix with zeros
{
    for(y=0; y<n; y++)
        magicsquare[x][y] = 0;
}

/*Beginning of the magic square algorithm*/
x = 0, y = n/2; //initialize algorithm at the middle column of the top row
for (counter = 1; counter <= square; counter++)    //magic square will contain the integers from 1 to n squared
{
    magicsquare[x][y] = counter;    //places current counter number at current position in the matrix or square
    x--;    //moves position diagonally up
    y++;    //and to the right

    /*If a move takes you above the top row in the jth column, move to the bottom of the jth column*/
    if(x<0)
        x = n - 1;
    /*If a move takes you outside to the right of the square in the ith row, move to the left side of the ith row*/
    else if(y==n)
        y = 0;
    /*If a move takes you to an already filled square or if you move out of the square at the upper right
    hand corner, move immediately below position of previous number*/
    else if((magicsquare[x][y] != 0) || (x<0 && y==n))
    {
        y--;    //move one space to the left back into the square
        x = x+2;    //move two spots down into the square and below previous number
    }
}
for(x=0; x<n; x++)
{
    for(y=0; y<n; y++)
        cout << setw(5) << magicsquare[x][y];
    cout << endl;
}

return 0;
}

Answer 1

我不能按照我的脑海中的逻辑来知道这是否真的会发生，但是在这段代码中：

if(x<0)
    x = n - 1;
/*If a move takes you outside to the right of the square in the ith row, move to the left side of the ith row*/
else if(y==n)
    y = 0;

如果两个条件都是真的，那么你将不会修复y，并且你将在矩阵的末尾运行下一次迭代。

另请注意，int magicsquare[n][n];是编译器扩展，C ++标准不支持，因为n不是编译时常量。你几乎肯定想要使用vector。

Answer 2

以下是非法的：

int magicsquare[n][n];

您是否忽略了错误，或者您使用的编译器是否完全没有错误？我建议你使用一个能在出错时提示你的IDE，这样你就可以很容易地看出你的错误了。请不要使用记事本来编写C ++，这太可怕了。

修正版：

int** magicsquare = new int*[n]; //creates a n by n matrix to set up magic square
for(int i = 0; i < n+1; ++i)
    magicsquare[i] = new int[n];

现在，与Mark B的提示一起，你马上就可以了。不要忘记使用删除方式清理magicsquare。

Answer 3

所以我对魔术方块一无所知。但我认为这是你想要实现的行为：

for (int counter = 1, x = 0, y = (n / 2); counter <= n * n; ++counter){
    magicsquare[x][y] = counter;    //places current counter number at current position in the matrix or square

    if (counter % n == 0){ //moves down into the square and below previous number
        x = (x + 1) % n;
    }
    else //moves position diagonally up and to the right
    {
        x = (x + n - 1) % n;
        y = (y + 1) % n;
    }
}

还有两点：

在我们使用Array Extensions技术规范之前，我认为您应该避免在代码中声明C99的运行时大小的数组。即使gcc允许它。您可能会考虑执行以下操作：vector<vector<int>> magicsquare(n, vector<int>(n));
这与Wikipedia's article所示的行为不符，但您可以通过调整索引的起始值和顺序来实现目标。

程序在某些输入上崩溃

3 个答案: