从1到n的正数中1出现的次数

题目：输入一个整数n，求从1到n这n个整数的十进制表示中1出现的次数。

例如输入12，从1到12这些整数中包含1 的数字有1，10，11和12，1一共出现了5次。

利用动态规划的思想，每次往前进行增加。代码如下：

#include<iostream>
using namespace std;


int Countdatanum(int n)
{
	int Data[10] = {0};
	int i = 0;
	int k = 10;
	int length;
	int temp;
	int sum = 0;
	int value = n;
	while(value != 0)
	{
		Data[i] = value % k;
		value /= 10;
		i++;
	}
	length = i;

	if(Data[0] > 0)
		Data[0] = 1;
	int pre = 1;
	sum += Data[0];
	for(i = 1; i < length; i++)
	{
		if(Data[i] > 1)
		{
			sum += pre * Data[i] + k;
		}
		else if(Data[i] == 1)
		{
			sum += pre * Data[i]; 
			sum += n % k + 1;
		}
		else
			Data[i] = 0;

		pre = pre * 10 + k;
		k *= 10;
	}
	
	return sum;
}

int NumberOf1(unsigned int n)
{
      int number = 0;
      while(n)
      {
            if(n % 10 == 1)
                  number ++;

            n = n / 10;
      }

      return number;
}

int NumberOf1BeforeBetween1AndN_Solution1(unsigned int n)
{
      int number = 0;

      // Find the number of 1 in each integer between 1 and n
      for(unsigned int i = 1; i <= n; ++ i)
            number += NumberOf1(i);

      return number;
}



int main()
{
	cout<<Countdatanum(203)<<endl;
	cout<<NumberOf1BeforeBetween1AndN_Solution1(203)<<endl;
	return 0;
}