Good Numbers

if we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number. 
You are required to count the number of good numbers in the range from A to B, inclusive.

Input

The first line has a number T (T <= 10000) , indicating the number of test cases. 
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 10¹⁸).

Output

For test case X, output "Case #X: " first, then output the number of good numbers in a single line.

Sample Input

21 101 20

Sample Output

Case #1: 0Case #2: 1          

Hint

The answer maybe very large, we recommend you to use long long instead of int.         分析：

发现有近似num / 10的规律，仔细观察后发现其实确实粗略来看是每10个出现一个和为10的倍数

因为对于一个数，例如1X，这时候个位只有10 - 1 = 9才满足，对于一个三位数例如12X，则由于1和2是确定的，所以在120 - 129这十个数的范围内（也就是个位0 - 9）只有10 - 1 - 2 = 7，即127满足，对于大一点的数，例如637859X，6 + 3 + 7 + 8 + 5 + 9 = 38，因为最后一位肯定只能为正（不可能为-8），所以只有最后一位为40 - 38 = 2，即6378592符合条件，所有的数都这么看，会发现每次高位确定了，其实在个位上只有0 - 9当中的一个数能够满足，这就印证了num / 10的规律
最后只要注意下右边界就行了，即
0 - 90 -> 90 / 10 = 9个
但是
0 - 91 ->90 / 10 + 1 = 10个（由于91这个边界多加了一个）
处理方法是从90 / 10 * 10 = 90 -->91暴力扫一下，看有没有多出来一个能够满足各位和是10的倍数的，发现91是，所以就在9的基础上加1，因为我们证明了个位0 - 9有且只有一个数满足，所以只要扫过去发现一个直接返回就行了
注意：0可以被10整除！！

import java.util.*;public  class Main{   static Scanner in = new Scanner(System.in);   static long get(long  num) { long n = num / 10 * 10;boolean flag = false;for (; n <= num; n++) {long tn = n, sum = 0;while (tn!=0) {sum += tn % 10;tn /= 10;}if (sum % 10 == 0) {flag = true;break;}}if (flag)return 1;elsereturn 0;}static long solve(long  num) {if (num < 0) return 0;if (num < 10) return 1;return num / 10 + get(num);}   public static void main(String args[]){         int k = in.nextInt();       int cas = 0;       while(k-->0) {       cas++;       long a = in.nextInt();       long b = in.nextInt();       long cnt = solve(b) - solve(a - 1);//-1就是为了把0算进去       System.out.println("Case #"+cas+": "+cnt);       }      }  }