hdu4722 Good Numbers

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

Sample Output

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

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#include <iostream>

#include <stdio.h>

#include <string.h>

using namespace std;

__int64 ff(__int64 m)

{

    if(m<0)

    return 0;

    __int64 temp=m/100,ans;

    __int64 i;

    ans=temp*10;

    for(i=temp*100;i<=m;i++)

    {

       __int64 sum=0,t=i;

       while(t)

       {

           sum+=t%10;

           t/=10;

       }

      if(sum%10==0)

      ans++;

    }

    return ans;

}

int main()

{

    int tcase ,tt=1;

    __int64 a,b;

    scanf("%d",&tcase);

    while(tcase--)

    {

        scanf("%I64d%I64d",&a,&b);

        printf("Case #%d: %I64d\n",tt++,ff(b)-ff(a-1));

    }

    return 0;

}