Digit sum-----The Preliminary Contest for ICPC Asia Shanghai 2019

A digit sum S_b(n)Sb​(n) is a sum of the base-bb digits of nn. Such as S_{10}(233) = 2 + 3 + 3 = 8S10​(233)=2+3+3=8, S_{2}(8)=1 + 0 + 0 = 1S2​(8)=1+0+0=1, S_{2}(7)=1 + 1 + 1 = 3S2​(7)=1+1+1=3.

Given NN and bb, you need to calculate \sum_{n=1}^{N} S_b(n)∑n=1N​Sb​(n).

InputFile

The first line of the input gives the number of test cases, TT. TT test cases follow. Each test case starts with a line containing two integers NN and bb.

1 \ leq T \ leq 100000 1 ≤ T ≤ 1 0 0 0 0 0

1 \ leq N \ leq 10 ^ 6 1 ≤ N ≤ 1 0 6

2 \ leq b \ leq 10 2 ≤ b ≤ 1 0

OutputFile

For each test case, output one line containing Case #x: y, where xx is the test case number (starting from 11) and yy is answer.

Sample input

2
10 10
8 2

Sample Output

Case #1: 46
Case #2: 13

#include <stdio.h>

int main()
{
    long long t, tot, n, b, a[15], x, num, sum, i, j, N, temp;
    for(i=0;i<13;i++)
    {
        a[i] = i;
    }
    scanf("%lld", &t);
    tot = 0;
    while(t--)
    {
        scanf("%lld %lld", &n, &b);
        temp = (b-1)*b/2;
        x = 1;
        sum = 0;
        while(n>=x)
        {
            sum = sum + temp * x * (n / (x * b));
            N = n % (x * b);
            num = 0;
            for(j=0;j+x<=N;j+=x)
            {
                if(num==b) num = 0;
                sum = sum + a[num] * x;
                num++;
            }
            if(num == b) whether = 0 ; 
            I = I + a [whether] * (n-j + 1 ); 
            x = b; 
        } 
        So ++ ; 
        printf ( " Case #% lid% lid \ n " , so many, I); 
    } 
    Return  0 ; 
}