标签:tar ace std namespace icpc integer 超时 efault 进制
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 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 yyis answer.
2 10 10 8 2
Case #1: 46 Case #2: 13
地址:https://nanti.jisuanke.com/t/41422
题意为:给出
t
N b
b为进制,求Sb(n)的加和,n从1~N 。比如样例2:S2(1~8),为S2(1)+S2(2)+...===1的二进制+2的二进制+3的二进制.....+8的二进制。而对于转化出的进制,由题意,S_{10}(233) = 2 + 3 + 3 = 8
S2?(8)=1+0+0=1。以此为规则。
暴力会超时,所以需要先进行预处理。这是一个累加,所以建立二维数组,i表示进制数:
1 2 3 4 5
i: 1
2
3
.
.
.
.
10
由于1的2~10进制均为1,所以有:
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!
Digit sum (第 44 届 ACM/ICPC 亚洲区域赛(上海)网络赛)进制预处理水题
标签:tar ace std namespace icpc integer 超时 efault 进制
原文地址:https://www.cnblogs.com/liyexin/p/11529714.html