CodeForces - 579A Raising Bacteria（水/water）

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.

 

题目大意：问要某一天的时候盒子里的细菌数量为n，需要你放入最少多少个细菌。（细菌每天一分为二）

解题思路：将n分解为多个2的几次幂相加，幂数尽量大。直接找就行。。

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=32;
int a[maxn];

void Init()
{
    a[0]=1;
    for(int i=1;i<=30;i++)
        a[i] = a[i-1]*2;
}

int main()
{
    int n;
    Init();
    while(scanf("%d",&n)!=EOF)
    {
        int num=0;
        int k;
        for(k=0;k<30;)
        {
            if(a[k]<=n&&a[k+1]>n)
            {
                //printf("%d %d\n",a[k],a[k+1]);
                num++;
                n -= a[k];
                if(n==0)
                {
                    break;
                }
                else
                {
                    k=0;
                }
            }
            else
                k++;
        }
        printf("%d\n",num);
    }
}