构建身份矩阵

时间:2012-12-17 20:01:53

标签: c++ algorithm

我试图解决算法问题,我是新手,我正在尝试在编程问题上练习很多。所以我想构造一个单位矩阵n * n。我提出了一个愚蠢的解决方案,适用于4 * 4矩阵,但它不适用于5 * 5。我知道,当我查看它时,它的奇怪解决方案和解决问题的方法非常简单。我需要知道我做错了什么才能让我学习,而且我的解决方案真的很愚蠢,以后我会在解决这些问题后做得更好吗?

#include <iostream>
#include <vector>
#include <sstream>
#include <iomanip>  // for setw, setfill

using namespace std;

int binary(int number);

int main()
{

    vector<vector<int> > matrix;

    cout<<"Please enter the size of the identity matrix"<<endl;
    int n;
    cin>>n;

    matrix.resize(n);
    for (int i=0; i<n;i++)
    {

        matrix[i].resize(n);
    }

    int steps = 1<<n-1;

    int bin  = binary(steps);
    ostringstream binString;

    binString <<bin;
    if(binString.str().size()<n)
    {
        std::string dest = binString.str();
        int nPaddings = n-binString.str().size();
        if (nPaddings==0) nPaddings=1;
        dest = std::string( nPaddings, '0').append( binString.str());
        binString.str("");
        binString<<dest;
    }
    for (int col = 0; col<n; col++)
    {
        if(col>=1)
        {
            steps= (int)steps/2;
            int bin = binary(steps);
            binString.str("");
            binString << bin;
            if(binString.str().size()<n)
            {
                std::string dest = binString.str();
                int nPaddings = n-steps;
                if (nPaddings==0) nPaddings=1;
                dest = std::string( nPaddings, '0').append( binString.str());
                binString.str("");
                binString<<dest;
            }
        }
        for (int row=0; row<n; row++)
        {
            matrix[col][row] =binString.str().at(row)-'0';
        }
    }


    return 0;
}

int binary(int number) {
    long rem,i=1,sum=0;
    do
    {
        rem=number%2;
        sum=sum + (i*rem);
        number=number/2;
        i=i*10;
    }while(number>0);

    return sum;
}

2 个答案:

答案 0 :(得分:3)

有一种更简单的方法。

首先,您应该使用指定的大小分配矩阵。然后,你知道只有对角线是1 s:

vector<vector<int> > matrix;
int n;

cout << "Please enter the size of the identity matrix" << endl;
cin >> n;

// Initialize the matrix as a n x n array of 0.
matrix = vector<vector<int> >(n, vector<int>(n,0));

// Set the diagonal to be 1s
for(unsigned int t = 0; t < n; t++)
    matrix[t][t] = 1;

您可以看到实时示例here

编辑:

您的错误来自这一行:

int nPaddings = n-steps;

实际上,您没有使用dest的大小来计算填充,这是不正确的。 See here,我添加了一些调试printfs来查看变量的状态。您可以看到nPaddings == -3,因此错误。

你的想法:

for each column
    get the representation of the column as a string
    set the i-th value of the column as the i-th character of the string

所以,这是一个使用你的想法的简单程序。在几个函数中分离代码有很大帮助。此外,std::ostringstreamstd::string在这里只是纯粹的过度杀伤。

#include <iostream>
#include <vector>
#include <iomanip>  // for setw, setfill

using namespace std;
std::string binStr(unsigned int exponent, unsigned int size);

int main()
{

    vector<vector<int> > matrix;

    cout<<"Please enter the size of the identity matrix"<<endl;
    int n;
    cin>>n;

// Initialize the matrix
    matrix.resize(n);
    for (int i=0; i<n;i++)
        matrix[i].resize(n);

// Fill the matrix
    for (int col = 0; col<n; col++)
    {
        std::string bin = binStr(n-col,n);
        for (int row=0; row<n; row++)
            matrix[col][row] = bin[row]-'0';
    }


// Print the matrix and return
    for(unsigned int y = 0; y < n; y++)
    {
        for(unsigned int x = 0; x < n; x++)
            cout << "\t" << matrix[y][x];
        cout << "\n";
    }
    return 0;
}

std::string binStr(unsigned int exponent, unsigned int size)
{
    // You do not need a string stream (which is like using a bazooka to kill a fly...)
    // Instead, just create a string of the required length
    // 'str' will contain the binary representation of 2^exponent
    std::string str(size,'0');
    if(exponent <= size && exponent > 0)
        str[size - exponent] = '1';
    return str;
}

您可以在行动here中看到它。

答案 1 :(得分:3)

#include <iostream>
#include <vector>
using namespace std;

vector<vector<int> > make_idty_matrix( int n )
{
    vector<vector<int> > idty( n, vector<int>( n, 0 ));
    for( int i = 0; i < n; ++i )
        idty[i][i] = 1;
    return idty;
}

int main()
{
    vector<vector<int> > matrix = make_idty_matrix( 5 );

    // your code here
    // ...

    return 0;
}