Raising Bacteria （Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] ）
Codeforces Round #320 (Div. 2) [Bayan Thanks-Round]的这个Raising Bacteria挺好玩的。
我也分享一下源码。
题目：
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.
Sample test(s)
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.
我的源码：
#include <iostream>
using namespace std;
int main()
{
int x;
while(cin>>x){
if(x>=1&&x<=109){
int answer=0;
while(x%2==0){x/=2;}
while(x){ //下面最后一次时x=1,，然后(x-1)%2->0,可以退出循环
if((x-1)%2==0){x=(x-1)/2;answer++;} //关键是找到answer++的地方，这个地方是放细菌的时候，而且我们不用关心天数。
else x/=2;
}
cout<<answer<<endl;
}
}
return 0;
}
（自己觉得这个解决方法还是挺好的，挺简洁的☺）