Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] A. Raising Bacteria【位运算/二进制拆分/细胞繁殖，每天倍增】

A. Raising Bacteria

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are a lover of bacteria. You want to raise some bacteria in a box.

Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly x bacteria in the box at some moment.

What is the minimum number of bacteria you need to put into the box across those days?

Input

The only line containing one integer x (1 ≤ x ≤ 109).

Output

The only line containing one integer: the answer.

Examples

 

input

5

output

2

input

8

output

1

Note

For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.

For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.

【题意】：在培养皿中，每个细胞每天会繁殖，数量*2 ，我们可以在任意天加入任意数量的细胞入培养皿中。 想要知道最少加入多少个细胞，可以使得有一天，培养皿中细胞的数量会恰好为x。

【分析】：x的二进制表示中1的个数即为答案．

原因是，每天晚上细胞数量翻倍，相当于左移１位,这时候二进制表示中1的数量不变。也就是说，二进制表示中的所有的1，一定都是添加进去的。而且也只有二进制表示中的1是添加进去的。所以二进制表示中1的数量，就是添加的细胞数。

【代码】：

#include<bits/stdc++.h>
using namespace std;
int n,m,cnt;

int main()
{
    while(cin>>n)
    {
        cnt=0; //多组输入内置清零
        while(n)
        {
            if(n%2==1)
                cnt++;
            n/=2;
        }
        cout<<cnt<<endl;
    }
    return 0;
}