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使用数学归纳法，代码如下：

/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;
}