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：

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #define LL _int64
 #include<algorithm>
 using namespace std;
 LL cal(LL n)
 {
     LL num,i,j,ans,sum;
      num=n/100;
     ans=num*10;
   for(i=num*100;i<=n;i++)
   {
        j=i;
       sum=0;
       while(j)
       {
           sum+=j%10;   //将各个位相加
           j/=10;
       }
       if(sum%10==0)
           ans++;
   }
  return ans;
 }
 void Init()
 {
     LL l,r;
     static int count=1;
     scanf("%I64d%I64d",&l,&r);
     LL temp=cal(r)-cal(l-1);
     printf("Case #%d: %I64d\n",count++,temp);
 }
 int main()
 {
   int t;
   scanf("%d",&t);
   while(t--)
   {
       Init();
   }
   return 0;
 }