标签：== turn ott div base int long one padding
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).
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
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.
2
10 10 8 2
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; }
Digit sum-----The Preliminary Contest for ICPC Asia Shanghai 2019
标签：== turn ott div base int long one padding
原文地址：https://www.cnblogs.com/0xiaoyu/p/11528861.html
