A digit sum Sb(n) is a sum of the base-bbb digits of nnn. Such as S10(233)=2+3+3=8 , S2(8)=1+0+0=1 S2(7)=1+1+1=3
Given N and b, you need to calculate ∑n=1NSb(n)
InputFile
The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line containing two integers N and b.
1≤T≤100000
1≤N≤10e6
2≤b≤10
OutputFile
For each test case, output one line containing Case #x: y, where xxx is the test case number (starting from 1) and y 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
题意:求一个数b进制,各个位数的和
一个个算的话必T,所以打表直接套之前求二进制位数的板子就过了,这次是2—10进制,简单总结下
for(int i=1; i<=maxn; i++) {
a2[i]=a2[i>>1]+i%2;
a3[i]=a3[i/3]+i%3;
a4[i]=a4[i/4]+i%4;
a5[i]=a5[i/5]+i%5;
a6[i]=a6[i/6]+i%6;
a7[i]=a7[i/7]+i%7;
a8[i]=a8[i/8]+i%8;
a9[i]=a9[i/9]+i%9;
a10[i]=a10[i/10]+i%10;
}
for(int i=1; i<=maxn; i++) {
a2[i]+=a2[i-1];
a3[i]+=a3[i-1];
a4[i]+=a4[i-1];
a5[i]+=a5[i-1];
a6[i]+=a6[i-1];
a7[i]+=a7[i-1];
a8[i]+=a8[i-1];
a9[i]+=a9[i-1];
a10[i]+=a10[i-1];
}
AC代码
代码太丑了,还是不写在博客里了,直接贴出来