传送门
579A. Raising Bacteria
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
题目描述
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?
输入描述
Input
The only line containing one integer x (1 ≤ x ≤ 109).
输出描述
Output
The only line containing one integer: the answer.
样例
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:细菌繁殖,一个一天时间可以变为两个,以此类推。求出加入多少个细菌可以确切知道在某天的细菌数量。
思路:跟2的倍数有关,用2进行迭代。
代码
#include <cstdio>#include <cstring>#include <iostream>using namespace std;int main(){int n;int sum=0;scanf(\"%d\",&n);while(n>1){if(n%2==1){sum++;}n=n/2;}printf(\"%d\\n\",sum+1);return 0;}