以十进制数的二进制格式计算 1 的个数

I am trying to find out the number of 1's in binary form of a large decimal number (decimal number can be as large as 1000000).

我试过这段代码:

while(sum>0)  
{  
    if(sum%2 != 0)  
    {  
        c++;   // counting number of ones  
    }  
    sum=sum/2;  
}  

我想要一个更快的算法，因为大十进制输入需要很长时间.请建议我一个有效的算法.

#include <bitset>
#include <iostream>
#include <climits>

size_t popcount(size_t n) {
    std::bitset<sizeof(size_t) * CHAR_BIT> b(n);
    return b.count();
}

int main() {
    std::cout << popcount(1000000);
}