2017年上海金马五校程序设计竞赛（网上资格赛） Problem J : Raising Bacteria

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本文介绍了一种通过计算细菌繁殖的数学模型来解决特定问题的方法。该问题要求找到最少需要放置多少细菌才能在某一天达到指定数量。通过预先计算2的幂次并使用贪心算法递减匹配目标值，实现了高效求解。

Description

You are a lover of bacteria so you want to raise some bacteria in a box. Initially, the box is empty. Every morning, you can put any number of bacteria into the box. Every 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

There are several test cases. Each case contains only one integer x (1≤x≤109) a line.

Output

For each case, output the only line containing one integer: the answer.

Sample Input

5
8

Sample Output

2
1

我是预估10^9的2的n次方大概在30，然后把他们算出来存到数组里，把给出的数找他最接近的那个减掉，在找最近的减掉，每次减掉加一个细胞就可以了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <math.h>
using namespace std;
int X[30];
int main()
{
	for(int i=0;i<30;i++)
	{
		X[i]=pow(2,i);
		
	}
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int count=0;
		for(int i=29;i>=0;i--)
		{
			if(n-X[i]>=0)
			{
				n-=X[i];
				count++;
				if(n<=0) break;
			 }  
		}
		printf("%d\n",count);
	}
}