四川省ACM竞赛（2013）---G-Good Numbers（规律/可DP）

G - Good Numbers

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4722

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. 
         

 


AC CODE:

#include <iostream>
#include <cstdio>
using namespace std;
long long cu(long long n)
{
    if(n<0) return 0;
    if(n<=10)   return 1;
    int s,more=0;
    long long i=n/10*10;
    long long m = n;
    for(; i<=n; i++)
    {
        m=i;
        s=0;
        while(m)
        {
            s+=m%10;
            m/=10;
        }
        if(s%10==0)more++;
    }
    return n/10+more;
}
int main()
{
    long long a,b;
    int cas=1;
    int T;
    cin>>T;
    while(T--)
    {
        cin>>a>>b;
        printf("Case #%d: ",cas++);
        cout<<cu(b)-cu(a-1)<<endl;
    }

    return 0;
}