Good Numbers

Problem Description

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

2
1 10
1 20

Sample Output

Case #1: 0 Case #2: 1

Hint

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

#include<cstdio>

__int64 sum(__int64 b)
{
    int flag=;
    __int64 s=b,sb;
    __int64 sum=;
    while(s!=)
    {
        sum+=s%;
        s/=;
    }
    sum=sum-b%;
    sum%=;
    if(sum== || sum+b%>=) flag=;
    else flag=;
    sb=b/-flag;
    return sb;
}

int main()
{
    __int64 a,b;
    int t,step=,k;
    scanf("%d",&t);
    while(t--)
    {
        k=;
        scanf("%I64d%I64d",&a,&b);
        if(a==) k=;
        printf("Case #%d: %I64d\n",step++,sum(b)-sum(a-)+k);
    }
    return ;
}

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