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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. Codeforces Beta Round #10 D. LCIS(DP&amp;LCIS)

    D. LCIS time limit per test 1 second memory limit per test 256 megabytes input standard input output ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. 深入理解Spring系列之八:常用的扩展接口

    转载 https://mp.weixin.qq.com/s/XfhZltSlTall8wKwV_7fKg Spring不仅提供了一个进行快速开发的基础框架,而且还提供了很多可扩展的接口,用于满足一些额 ...

  2. gpk-update-icon占用CPU及清除【原创】

    发现服务器有个gpk-update-icon一直占用CPU进程 网上查看相关信息比较少. gpk-update-icon是gnome的更新图标进程 俩种处理方法: 1.杀掉gpk-update-ico ...

  3. PCA算法和SVD

    如果矩阵对某一个向量或某些向量只发生伸缩变换,不对这些向量产生旋转的效果,那么这些向量就称为这个矩阵的特征向量,伸缩的比例就是特征值.这里可以将特征值为负,特征向量旋转180度,也可看成方向不变,伸缩 ...

  4. 在数据库中(Oracle),根据时间查询数据: to_date()和to_char()函数

    1. to_date() 函数 1.1 格式 to_date("要转换的字符串","转换的格式")   //两个参数的格式必须匹配,否则会报错.即按照第二个参数 ...

  5. Python股票信息抓取~

    本来想把股票的涨跌抓取出来,用汇通网的股票为例,就找了国际外汇为例. 页面里有xhr请求,并且每个xhr的url请求的 http://api.q.fx678.com/history.php?symbo ...

  6. ios测试apk

    最近apk在ios上面测试总是会遇到奇奇怪怪的问题,现在是两个项目要集成在一个apk中所以将两个项目运行之后都是编译成了.a文件,然后在两个.a文件中都设置了两个意义相同变量名相同的全局变量(标识当前 ...

  7. gentoo emerge unable to sync

    gentoo emerge unable to sync Author: Tubo After setting SYNC to customized URL: SYNC="rsync://m ...

  8. AC日记——[CQOI2009]DANCE跳舞 洛谷 P3153

    [CQOI2009]DANCE跳舞 思路: 二分+最大流: 代码: #include <cstdio> #include <cstring> #include <iost ...

  9. qt调用仪器驱动库dll实现程控

    在<使用qt+visa实现程控>中实现了qt调用visa库的简单Demo本文将尝试使用qt调用仪器驱动库来实现对仪器仪表的程控 开发环境 系统: windows 10 环境: qt 5.8 ...

  10. day3修改配置文件

    有如下配置文件,在指定文件位置添加一条新的记录: global log 127.0.0.1 local2 daemon maxconn log 127.0.0.1 local2 info defaul ...