Codeforces Beta Round #10 C. Digital Root 数学
C. Digital Root
题目连接:
http://www.codeforces.com/contest/10/problem/C
Description
Not long ago Billy came across such a problem, where there were given three natural numbers A, B and C from the range [1, N], and it was asked to check whether the equation AB = C is correct. Recently Billy studied the concept of a digital root of a number. We should remind you that a digital root d(x) of the number x is the sum s(x) of all the digits of this number, if s(x) ≤ 9, otherwise it is d(s(x)). For example, a digital root of the number 6543 is calculated as follows: d(6543) = d(6 + 5 + 4 + 3) = d(18) = 9. Billy has counted that the digital root of a product of numbers is equal to the digital root of the product of the factors' digital roots, i.e. d(xy) = d(d(x)d(y)). And the following solution to the problem came to his mind: to calculate the digital roots and check if this condition is met. However, Billy has doubts that this condition is sufficient. That's why he asks you to find out the amount of test examples for the given problem such that the algorithm proposed by Billy makes mistakes.
Input
The first line contains the only number N (1 ≤ N ≤ 106).
Output
Output one number — the amount of required A, B and C from the range [1, N].
Sample Input
4
Sample Output
2
Hint
题意
问你[1,n]中有多少 AB!=C,但是D(A)D(B)=D(C)的
D(A)是数根的意思,翻译过来就是这个数%9
题解:
容斥做,首先把所有的D(A)D(B)=D(C)的计算过来
然后减去AB==C且D(A)D(B)=D(C)的,由于显然AB=C,那么D(A)D(B)=D(C)
所以我们只需要减去AB=C的就好了,我们暴力枚举A,看B的个数有n/A个
然后莽一波……
代码
#include<bits/stdc++.h>
using namespace std;
int a[10];
int main()
{
int n;long long ans = 0;
scanf("%d",&n);
for(int i=1;i<=n;i++)a[i%9]++,ans-=n/i;
for(int i=0;i<9;i++)for(int j=0;j<9;j++)ans+=1ll*a[i]*a[j]*a[i*j%9];
cout<<ans<<endl;
}
Codeforces Beta Round #10 C. Digital Root 数学的更多相关文章
- Codeforces Beta Round #10 D. LCIS
题目链接: http://www.codeforces.com/contest/10/problem/D D. LCIS time limit per test:1 secondmemory limi ...
- Codeforces Beta Round #10 D. LCIS 动态规划
D. LCIS 题目连接: http://www.codeforces.com/contest/10/problem/D Description This problem differs from o ...
- Codeforces Beta Round #10 B. Cinema Cashier 暴力
B. Cinema Cashier 题目连接: http://www.codeforces.com/contest/10/problem/B Description All cinema halls ...
- Codeforces Beta Round #10 A. Power Consumption Calculation 水题
A. Power Consumption Calculation 题目连接: http://www.codeforces.com/contest/10/problem/A Description To ...
- Codeforces Beta Round #11 B. Jumping Jack 数学
B. Jumping Jack 题目连接: http://www.codeforces.com/contest/11/problem/B Description Jack is working on ...
- Codeforces Beta Round #10 D. LCIS(DP&LCIS)
D. LCIS time limit per test 1 second memory limit per test 256 megabytes input standard input output ...
- Codeforces Beta Round #10 B. Cinema Cashier (树状数组)
题目大意: n波人去k*k的电影院看电影. 要尽量往中间坐,往前坐. 直接枚举,贪心,能坐就坐,坐在离中心近期的地方. #include <cstdio> #include <ios ...
- Codeforces Beta Round #80 (Div. 2 Only)【ABCD】
Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一 ...
- Codeforces Beta Round #13 C. Sequence (DP)
题目大意 给一个数列,长度不超过 5000,每次可以将其中的一个数加 1 或者减 1,问,最少需要多少次操作,才能使得这个数列单调不降 数列中每个数为 -109-109 中的一个数 做法分析 先这样考 ...
随机推荐
- css的背景图片background
1.使用背景图片的标签定设置宽高,没有设置的话,也需要用内容来撑开标签. 2.如果对同一个标签分开设置背景图片和颜色,背景颜色一定要写在背景图片后面,不然会被覆盖 <!DOCTYPE html& ...
- u-boot中的Makefile
在windos下,pc机上电之后,BIOS会初始化硬件配置,为内核传递参数,引导操作系统启动,并且识别C盘.D盘.等整个操作系统启动起来之后,才可以运行应用程序比如QQ.QQ音影.同理,在嵌入式Lin ...
- cgi与html相互调用
html中调用cgi.<form action="/cgi-bin/mult.cgi" method="get" target="_blank& ...
- C++随笔(2)
在牛客网上刷题,遇到的一些需要注意的题 1.这题需要注意的是strcpy复制的时候什么时候停止 2.这题是关于strlen的,它不统计‘\0',但复制的时候仍会复制. 3.这题是写strcpy函数的, ...
- Effective C++笔记(四):设计与声明
参考:http://www.cnblogs.com/ronny/p/3747186.html 条款18:让接口容易被正确使用,不易被误用 1,好的接口很容易被正确使用,不容易被误用.你应该在你的所有接 ...
- java图片转byte转string
第一种:原始乱码: public static void main(String[] args) throws IOException { File imgFile = new File(" ...
- 在eclipse中使用Maven3(笔记二)
笔记本二 在Eclipse 中使用Maven 第一节:m2eclipse 插件安装 打开Eclipse,点击菜单Help - > Install New Software 点击Add 按钮N ...
- golang-goroutine和channel
goroutine 在go语言中,每一个并发的执行单元叫做一个goroutine 这里说到并发,所以先解释一下并发和并行的概念: 并发:逻辑上具备同时处理多个任务的能力 并行:物理上在同一时刻执行多个 ...
- DB2和Oracle中Date比较
- 如何去除decimal后面的零?
如何去除decimal后面的零? 1.260000m.ToString("G29") 不显示科学记数法? decimal.Parse("0.0000001",S ...