概述
| Time Limit: 1000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
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
5
2
8
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.
x 的二进制表示中所有的 1 的个数就是答案，因为每天晚上细菌的数量会加倍，表现在二进制上面就是 1<<2，1的个数没有变，只是位置左移了，所以二进制表示中所有的1 都是添加进去的。 1 的个数也就是answer
#include <iostream>
using namespace std;
int main()
{
int x,i,j,ans = 0;
cin>>x;
while(x){
if(x%2==1)
ans ++;
x=x/2;
}
cout<<ans<<endl;
return 0;
}
最后
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