二进制表示中数字的偶数和奇数位的1的数量是多少？(What is the number of 1's at even and odd places of a number in binary representation?)

 无论是按位还是64位范围内的整数函数，它们都是更快的实现方式。除了我实现的那个。 
 
/* 
Find F(i)=abs(a(i)-b(i))
a(i)=number of 1's in even position 
b(i)=number of 1's in odd position 
for an integer i, where i fits in 64-bit
*/
//function calculate the above equation
//returns the answer
long long int F(long long int k)
{
    //size of array is taken for 64-bit number
    int a[64]={0},i,a,b;
    long long int m;
    m=k;
    //convert long long int into binary 
    for(i=63;i>-1;i--)
    {
        if(m==1||m==0)
        {
            a[i]=m;
            break;       //exit the for loop
        }
        a[i]=m%2;        //storing bit by bit
        m/=2;           
    }
    // initialized with a value of zero
    a=0;
    b=0;
    //find first bit having 1
    int f;
    for(i=0;i<64;i++)
    {
        if(a[i]==1)
        {
            f=i;
            break;
        }
    }
    //calculating the number of 1's in even and odd positions
    for(i=f;i<64;i++)
    {
        if(a[i]==1)
        {
            if((63-f)%2==0)
            {
                a++;          //1's in even positions
            }
            else
            {
                b++;          //1's in odd positions
            }
        }
    }

    //return the answer
    return abs(a-b);
}
 
 所以基本上我要做的是通过使用mod 2的简单方法转换其二进制表示中的整数。然后执行任务以从左到右找到其二进制表示中的第一个1，并且我们的指针在第一个数。 现在使用第一个索引计算奇数和偶数位置的1的数量。最后返回总偶数和奇数位置1的绝对差值。 

Is their is any faster way to implement it whether it is bitwise or any function for an integer in the range of 64-bit.Except the one I've implemented.
 
/* 
Find F(i)=abs(a(i)-b(i))
a(i)=number of 1's in even position 
b(i)=number of 1's in odd position 
for an integer i, where i fits in 64-bit
*/
//function calculate the above equation
//returns the answer
long long int F(long long int k)
{
    //size of array is taken for 64-bit number
    int a[64]={0},i,a,b;
    long long int m;
    m=k;
    //convert long long int into binary 
    for(i=63;i>-1;i--)
    {
        if(m==1||m==0)
        {
            a[i]=m;
            break;       //exit the for loop
        }
        a[i]=m%2;        //storing bit by bit
        m/=2;           
    }
    // initialized with a value of zero
    a=0;
    b=0;
    //find first bit having 1
    int f;
    for(i=0;i<64;i++)
    {
        if(a[i]==1)
        {
            f=i;
            break;
        }
    }
    //calculating the number of 1's in even and odd positions
    for(i=f;i<64;i++)
    {
        if(a[i]==1)
        {
            if((63-f)%2==0)
            {
                a++;          //1's in even positions
            }
            else
            {
                b++;          //1's in odd positions
            }
        }
    }

    //return the answer
    return abs(a-b);
}
 
So basically what I am trying to do is to convert the integer in its binary representation by simple method of using mod 2. Then a task is performed to find the first 1 in its binary representation for left to right and our pointer is on the first number. Now count the number of 1's in odd and even position using the index of first 1.Finally return the absolute difference of total even and odd place 1's.

原文：https://stackoverflow.com/questions/25814786