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 1000001≤T≤100000

1 \leq N \leq 10^61≤N≤106

2 \leq b \leq 102≤b≤10

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.

样例输入

2
10 10
8 2

样例输出

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) num=0;
            sum = sum + a[num]*(N-j+1);
            x *= b;
        }
        tot++;
        printf("Case #%lld: %lld\n", tot, sum);
    }
    return 0;
}