编程之美中1的个数的问题

//采用最直接的方法，时间复杂度为O（N*logN）；
/*#include <iostream>
#include<typeinfo>
#include<specstrings.h>
#include<string>
using namespace std;
int num_one(int n)
{
	int count=0;
	while(n)
	{
		count+=(n%10==1)?1:0;
		n/=10;
	}
	return count;
}
int  core(int num)
{
	int i=1;
	int count=0;
	for(;i<=num;++i)
	{
		count+=num_one(i);
	}
	return count;
}
  int main()
  {
	  cout<<core(10)<<endl;
	  system("pause");
	  return 0;
  }*/
//下面采取对每位进行分析的方法对来求的1的个数
#include<iostream>
using namespace std;
int num_one(int n)
{
	int count=0;
	int low=0;
	int cur=0;
	int high=0;
	int factor=1;
	while(n/factor!=0)
	{
		low=n-(n/factor)*factor;
		cur=(n/factor)%10;
		high=n/(factor*10);
		switch(cur)
		{
		case 0:
			count+=high*factor;
			break;
		case 1:
			count+=high*factor+low+1;
			break;
		default :
			count+=(high+1)*factor;
			break;
		}
		factor*=10;

	}
	return count;
}
int main()
{
	cout<<num_one(5)<<endl;
	system("pause");
	return 0;
}