优化转换算法

我最近一直在做一本我一直在读的书中的练习。任务是编写一个程序，打印出1-256之间的所有数字，分别是二进制、八进制和十六进制。我们只应该使用到目前为止在书中所学到的方法，这意味着只使用for, while和do. while循环，if和else if语句，将整数转换为等效的ASCII和一些更基本的东西(例如cmath和iomanip)。

所以经过一些工作，这是我的结果。然而，它是混乱的、不优雅的和模糊的。有没有人有任何提高代码效率(或优雅)的建议?P)和性能?

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
int decimalValue, binaryValue, octalValue, hexadecimalValue, numberOfDigits;
cout << "DecimalttBinaryttOctalttHexadecimalnn";
for (int i = 1; i <= 256; i++)
{
    binaryValue = 0;
    octalValue = 0;
    hexadecimalValue = 0;
    if (i != 0) 
    {
    int x, j, e, c, r = i, tempBinary, powOfTwo, tempOctal, tempDecimal;
    for (j = 0; j <=8; j++) //Starts to convert to binary equivalent
    {
        x = pow(2.0, j);
        if (x == i)
        {
              powOfTwo = 1;
              binaryValue = pow(10.0, j);
              break;
        }
        else if (x > i)
        {
              powOfTwo = 0;
              x /= 2;
              break;
        }
    }
    if (powOfTwo == 0)
    {
    for (int k = j-1; k >= 0; k--)
    {
        if ((r-x)>=0)
        {
           r -= x;
           tempBinary = pow(10.0, k);
           x /= 2;
        }
        else if ((r-x)<0)
        {
           tempBinary = 0;
           x /= 2;
        }
        binaryValue += tempBinary;
    }
    } //Finished converting
    int counter = ceil(log10(binaryValue+1)); //Starts on octal equivalent
    int iter;
    if (counter%3 == 0)
    {
       iter = counter/3;
    }
    else if (counter%3 != 0)
    {
       iter = (counter/3)+1; 
    }
    c = binaryValue;
    for (int h = 0; h < iter; h++)
    {
        tempOctal = c%1000;
        int count = ceil(log10(tempOctal+1));
        tempDecimal = 0;
        for (int counterr = 0; counterr < count; counterr++)
        {
            if (tempOctal%10 != 0)
            {
                 e = pow(2.0, counterr);
                 tempDecimal += e;
            }
            tempOctal /= 10;
        }
        octalValue += (tempDecimal * pow(10.0, h));
        c /= 1000;
    }//Finished Octal conversion
    cout << i << "tt" << binaryValue << setw(21-counter) << octalValue << "tt";
    int c1, tempHex, tempDecimal1, e1, powOf;
    char letter;
    if (counter%4 == 0)//Hexadecimal equivalent
    {
       iter = counter/4;
    }
    else if (counter%4 != 0)
    {
       iter = (counter/4)+1;
    }
    c1 = binaryValue;
    for (int h = 0, g = iter-1; h < iter; h++, g--)
    {
        powOf = g*4;
        if (h == 0)
        {
              tempHex = c1 / pow(10.0, powOf);
        }
        else if (h > 0)
        {
             tempHex = c1 / pow(10.0, powOf);
             tempHex %= 10000;
        }
        int count = ceil(log10(tempHex+1));
        tempDecimal1 = 0;
        for (int counterr = 0; counterr < count; counterr++)
        {
            if (tempHex%10 != 0)
            {
                 e1 = pow(2.0, counterr);
                 tempDecimal1 += e1;
            }
            tempHex /= 10;
        }
        if (tempDecimal1 <= 9)
        {
        cout << tempDecimal1;
        }
        else if (tempDecimal1 > 9)
        {
        cout << char(tempDecimal1+55); //ASCII's numerical value for A is 65. Since 10-15 are supposed to be letters you just add 55
        }
    }
    cout << endl;
    }
}
system("pause");
return 0;
}