HDUOJ----Good Numbers

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 673    Accepted Submission(s): 242

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.

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

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代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #define LL _int64
 6 #include<algorithm>
 7 using namespace std;
 8 LL cal(LL n)
 9 {
10     LL num,i,j,ans,sum;
11      num=n/100;
12     ans=num*10;
13   for(i=num*100;i<=n;i++)
14   {
15        j=i;
16       sum=0;
17       while(j)
18       {
19           sum+=j%10;   //将各个位相加
20           j/=10;
21       }
22       if(sum%10==0)
23           ans++;
24   }
25  return ans;
26 }
27 void Init()
28 {
29     LL l,r;
30     static int count=1;
31     scanf("%I64d%I64d",&l,&r);
32     LL temp=cal(r)-cal(l-1);
33     printf("Case #%d: %I64d\n",count++,temp);
34 }
35 int main()
36 {
37   int t;
38   scanf("%d",&t);
39   while(t--)
40   {
41       Init();
42   }
43   return 0;
44 }