HDU-4722-Good Numbers（找规律）

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Good Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5185 Accepted Submission(s): 1638

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 
#include 
#include
typedef long long ll;
ll solve(ll a)
{
    ll sum=a/10;
    ll x=a/10*10;
    for(ll i = x; i <= a; i ++)
    {
        ll num=0;
        ll y=i;
        while(y)
        {
            num=(y%10)+num;
            y/=10;
        }
        if(num%10==0)
            return sum+1;
    }
    return sum;
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int k=1; k<=t; k++)
    {
        ll x,y;
        scanf("%lld%lld",&x,&y);
        x--;
        ll x1,x2;
        x1=solve(x);
        if(x<0)
            x1=0;
        x2=solve(y);
        ll ans=abs(x2-x1);
        printf("Case #%d: %lld\n",k,ans);
    }
    return 0;
}