Digit sum
33.57% 2000ms 131072K
A digit sum S_b(n)S
b
​
(n) is a sum of the base-bb digits of nn. Such as S_{10}(233) = 2 + 3 + 3 = 8S
10
​
(233)=2+3+3=8, S_{2}(8)=1 + 0 + 0 = 1S
2
​
(8)=1+0+0=1, S_{2}(7)=1 + 1 + 1 = 3S
2
​
(7)=1+1+1=3.

Given NN and bb, you need to calculate \sum_{n=1}^{N} S_b(n)∑
n=1
N
​
S
b
​
(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≤10
6

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 复制
2
10 10
8 2
样例输出 复制
Case #1: 46
Case #2: 13

题目大意：
先输入一个整数$T$，代表有$T$组测试数据，下面$T$行每行输入两个整数$n,b$，先假设函数$f(x,b)$为$x$转换为$b$进制后各数位的和，例如$f(8,2)=1,f(10,10)=1$，求${\sum }_{i=1}^{n}f(i,b)$

解题思路：
此题无需想的太复杂，因为$1\le n\le 1e6$，且$2\le b\le 10$，所以可以将$1e6$以内所有的情况全都暴力打出来，最后$O(1)$输出即可。

代码:

//#pragma GCC optimize(3,"Ofast","inline")
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
int get_sum(int n,int b) {int sum=0;while(n) {sum+=n%b;n=n/b;}return sum;
}
int sum[1000100][11];
int main() 
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);#endif//ios::sync_with_stdio(0),cin.tie(0);rep(i,1,1000000) {rep(j,2,10) {sum[i][j]=sum[i-1][j]+get_sum(i,j);}}int T;scanf("%d",&T);rep(t,1,T) {int n,b;scanf("%d %d",&n,&b);printf("Case #%d: %d\n",t,sum[n][b]);}return 0;
}