Codeforces 579A. Raising Bacteria(位运算)

最新推荐文章于 2019-08-14 13:58:00 发布

原创最新推荐文章于 2019-08-14 13:58:00 发布 · 553 阅读

0 ·

CC 4.0 BY-SA版权

杂题同时被 3 个专栏收录

42 篇文章

订阅专栏

其他

20 篇文章

订阅专栏

Codeforces

8 篇文章

订阅专栏

这篇博客探讨了如何在细菌每晚分裂的情况下，通过最小数量的初始细菌在某时刻达到指定数量。问题转化为寻找目标数量的二进制表示中1的个数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 xbacteria 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

solution：

找一下n的二进制中1的个数

#include<cstdio>
using namespace std;
int main()
{
	int n,ans=0;
	scanf("%d", &n);
	while (n)
	{
		if (n & 1)ans++;
		n >>= 1;
	}
	printf("%d\n", ans);
	return 0;
}