Question

我想要的是接受一个整数，并在方法中将其转换为二进制，接近下面代码中显示的方法。但是，我想将数字转换为二进制，就像它在小数点后面一样。因此，如果我输入625作为输入，那么我希望它转换为101。

鉴于代码：

#include <iostream>
using namespace std;

int decTobinary(int);

int main(){
    cout << "Enter a number = ";
    int num;
    cin >> num; 
    int answer = decTobinary(num);
    cout << "answer: " << answer << endl"
    return 0;
}
int decTobinary(int x) {
    if (x==0)
    return 0;
    return 10 * decTobinary(x/2) + x % 2;
}

Answer 1

我把它一起砍掉了所以它并不漂亮，但它提供了你想要的输出：

#include <iostream>

float intToDecimalPoint(float f)
{
    if((f / 10.0f) <= 1.0f)
        return f/10.0f;
    else
        return intToDecimalPoint(f/10.0f);
}

void decTobinary(int x)
{
    if (x==0)
        return;

    float decimalPoint = intToDecimalPoint(float(x));
    if(decimalPoint*2.0f >= 1.0f)
    {
        std::cout << 1;
        int newX = int((decimalPoint*2.0f - 1.0f)*(float(x)/decimalPoint));
        return decTobinary(newX);
    }
    else
    {
        std::cout << 0;
        decTobinary(x*2);
    }
}

int main()
{
    std::cout << "Enter a number = ";
    int num;
    std::cin >> num; 
    std::cout << "answer: ";
    decTobinary(num);
    std::cout << std::endl;


    std::cin.get();
    std::cin.get();
    return 0;
}

Algo来自here。由于浮点不精确，我的解决方案并不完美，但它适用于大多数情况？

Answer 2

你可以尝试

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

int GetNumberOfDigits (int i)
{
    return i > 0 ? (int) log10 ((double) i) + 1 : 1;
};

void function(int number, vector<int> &convert)    {
    double remainder = number/pow(10,GetNumberOfDigits(number));
    do
    {
        remainder = remainder*2;
        convert.push_back(int(remainder));
        remainder = remainder - int(remainder);
    }
    while(remainder != 0);
};

int main()  {
    vector<int> solution;
    int n;
    cout<<"Enter number: ";
    cin>>n;
    function (n, solution);
    for(int index = 0; index < solution.size(); index++) std::cout<<solution[index];
    return 1;
}

Answer 3

Archive for the ‘C Programming’ Category http://www.programmingclub.in/category/c-programming/

#include<stdio.h>

#include<conio.h>

void main()

{

  int a[10],n,i,j=0;

  printf(“Enter the Number: “);

  scanf(“%d”,&n);

  while(n>0)

  {

             i=n%2;

             a[j++]=i;

             n=n/2;

            }

  j–;

  printf(“\nBinary form is: \n”);

  while(j>=0)

  {

             printf(“%d”,a[j]);

             j–;

            }

            getch();

  }

C ++，代码写小数点后的整数的二进制形式

3 个答案: