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 <= 10 ¹⁸).

 

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 }

View Code