编程之美---1的个数 C++实现

方法1

#include "stdafx.h"
#include 
#include 
using namespace std;

int count(int n){
	int result=0;
	while(n!=0){
		if(n%10==1)
			result++;
		n/=10;
		}
	return result;
	}

int total(int n){
	int sum=0;
	for(int i=1;i<=n;i++)
		sum+=count(i);
	return sum;
}

int _tmain(int argc, _TCHAR* argv[])
{
	time_t t_start,t_end;
	t_start=time(NULL);
	cout<

方法2

#include "stdafx.h"
#include 
#include 
using namespace std;

int total(int n){
	int count=0;
	int cur=0,low=0,high=0;
	int factor=1;

	while(n!=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;
		n/=factor;
		}
	return count;
	
	}


int _tmain(int argc, _TCHAR* argv[])
{
	time_t t_start,t_end;
	t_start=time(NULL);
	cout<