给定一个十进制n，从1开始，到n的所有整数，求出现1的个数

/统计出现1的个数，使用数学归纳法，得出的规律如下：
//分别计算每一位可能出现的1的次数，然后把每一位的次数相加
//思想：1、若当前位为0，则出现的次数仅由其更高位的数字决定，值为：更高位数字*当前位所在的位数
//2、若当前位为1，则出现的次数有更高位和其低位决定，值为：更高位数字*当前位所在的位数+更低位数字+1；
//3、若当前位>1,则出现的次数由更高位决定，值为：(更高位数字+1)*当前位所在的位数
long long Sum1(long long number)
{
long long count = 0;
long long factor = 1;//除数因子，用于截取出当前位，更低位，更高位,直到number被分解为0
long long higher_num = 0;
long long lower_num = 0;
long long curr_num = 0;
while(number / factor != 0)
{
lower_num = number - (number/factor)*factor;
curr_num = (number/factor)%10;
higher_num = number/(factor*10);
switch(curr_num)
{
case 0:
count += higher_num * factor;
break;
case 1:
count += higher_num * factor + lower_num +1;
break;
default:
count += (higher_num + 1)*factor;
break;
}
factor *= 10;
}
return count;
}