C++实现十进制数转换为二进制数的数学算法

 eg:13(10)----1101(2):
  1. remainder=n%2=1; n=n/2=6; y=y+remainder*flag=0+1*1=1;flag=flag*10=10;
  2. remainder=n%2=0; n=n/2=3; y=y+reminder*flag=1+0*10=1;
flag=flag*10=100;
  4. remainder=n%2=1; n=n/2=1; y=y+reminder*flag=1+1*100=101;
flag=flag*10=1000;
  5. remainder=n%2=1; n=n/2=0;y=y+reminder*flag=101+1*1000=1101;
flag=flag*10=10000;
此时n=0，if(n==0)成立，停止循环，输出y。

int transfer(int n)
{
    int flag=1,y=0,remainder; 
    while(1)
    {
        remainder=n%2;
        n/=2;
        y+= remainder*flag;
       flag*=10;
        if(n==0)
        {
            break;
        }
    }
    return y;
}

#include <cstdio>
#include<iOStream>
using namespace std;
int a[1000001];
int main()
{
    int transfer(int n);
    int n,sum=0,y;
    scanf("%d",&n);
    y=transfer(n);   
    printf("%d\n",y);
    return 0;
}
int transfer(int n)
{
    int flag=1,y=0,remainder;
    while(1)
    {
        remainder=x%2;
        n/=2;
        y+=remainder*flag;
        flag*=10;
        if(n==0)
        {
            break;
        }
    }
    return y;
}