Digit sum 打表签到（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 yyis answer.

样例输入复制

2
10 10
8 2

样例输出复制

Case #1: 46
Case #2: 13

---

暴力打表，想运行时间短一些的话，把结果打出来储存的话，会很快，不然的话，时间有点长。建议scanf,cin的话会T。

lli a[11][1000000+5];
int main()
{
    TLE;
    int n,k,t;
    lli ans,cnt; 
    for(int i=2;i<=10;i++)
    {
        for(int j=1;j<=1e6;j++)
        {
            ans = j; cnt = 0;
            while(ans)
            {
                cnt+=ans%i;
                ans/=i;
            } 
            a[i][j]=a[i][j-1]+cnt;
        } 
    }
    //cin>>t;
    read(t);
    for(int i=1;i<=t;i++)
    {
        read2(n,k);
        //cin>>n>>k;
        printf("Case #%d: %lld
",i,a[k][n]);
        //cout<<"Case #"<<i<<": "<<a[k][n]<<endl;
    }
    ok;
}