Digit sum （第 44 届 ACM/ICPC 亚洲区域赛（上海）网络赛）进制预处理水题

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

Given $\sum_{n=1}^{N} S_b(n)∑n=1NSb(n).$

InputFile

The first line of the input gives the number of test cases,

$\leq T \leq 1000001≤T≤100000$

$\leq N \leq 10^61≤N≤106$

$\leq b \leq 102≤b≤10$

OutputFile

For each test case, output one line containing Case #x: y, where

样例输入

2 10 10 8 2

样例输出

Case #1: 46 Case

    for(int i=1;i<=10;i++)     {         a[i][1]=1;     }

　　　　　　然后在j中，2=1+2，3=1+2+3，就是个累加，得到方程a[i][j]=a[i][j-1]+ac(j,i);ac里获取，i进制的j的各个位数相加。

    for(int i=2;i<=10;i++)     {         for(int j=2;j<=maxn;j++)         {             a[i][j]=a[i][j-1]+ac(j,i);         }     }

总的代码为：：

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm>  const int maxn=1e6+10;  using namespace std; typedef long long ll; ll a[11][maxn]; int ac(ll n,ll b) {     ll sum=0;     while(n)     {         sum+=n%b;         n=n/b;     }     return sum; } int main() {     for(int i=1;i<=10;i++)     {         a[i][1]=1;     }     for(int i=2;i<=10;i++)     {         for(int j=2;j<=maxn;j++)         {             a[i][j]=a[i][j-1]+ac(j,i);         }     }     ll t;     scanf("%lld",&t);     int ak=1;     while(t--)     {         ll n,b;         scanf("%lld%lld",&n,&b);         printf("Case #%d: %lld\n",ak++,a[b][n]);     } }

　　　　　　over!

来源：博客园

作者：liyeixn

链接：https://www.cnblogs.com/liyexin/p/11529714.html

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acm