codeforces #320 div2A Raising Bacteria

G - Raising Bacteria
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
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Description

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.

Sample Input

Input

5

Output

2

Input

8

Output

1

Hint

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.

思路：这个题奇数个是不可能通过细菌分裂得来的只能我们添加，因为没规定天数，所以偶数就让他们分裂，其实就是求给定数转化为二进制中1的个数

#include<stdio.h>

int main()

{ int x,sum=0;

scanf("%d",&x);

while(x>0)

{ if(x%2==1)

sum++;

x/=2;

}

printf("%d",sum);

return 0;

}