数学场,做到怀疑人生系列

C - Flip,Flip, and Flip......


Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up.

We will perform the following operation once for each square contains a card:

  • For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square.

It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations.

Constraints

  • 1≤N,M≤109
  • All input values are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of cards that face down after all the operations.


Sample Input 1

Copy
2 2

Sample Output 1

Copy
0

We will flip every card in any of the four operations. Thus, after all the operations, all cards face up.


Sample Input 2

Copy
1 7

Sample Output 2

Copy
5

After all the operations, all cards except at both ends face down.


Sample Input 3

Copy
314 1592

Sample Output 3

Copy
496080

这个可以找规律,314×5×k的末尾是0 ,然后不断找下去,竟然是这个数-2相乘

#include<bits/stdc++.h>
using namespace std;
long long a,b;
int main(){
cin>>a>>b;
cout<<abs((a-)*(b-));
}

D - Remainder Reminder


Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

Takahashi had a pair of two positive integers not exceeding N(a,b), which he has forgotten. He remembers that the remainder of a divided by b was greater than or equal to K. Find the number of possible pairs that he may have had.

Constraints

  • 1≤N≤105
  • 0≤KN−1
  • All input values are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of possible pairs that he may have had.


Sample Input 1

Copy
5 2

Sample Output 1

Copy
7

There are seven possible pairs: (2,3),(5,3),(2,4),(3,4),(2,5),(3,5) and (4,5).


Sample Input 2

Copy
10 0

Sample Output 2

Copy
100

Sample Input 3

Copy
31415 9265

Sample Output 3

Copy
287927211

这个题目骚啊,去暴力统计每个值对应的方案

#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,k;
cin>>n>>k;
long long ans=;
for(int i=k+; i<=n; i++)
ans+=(n/i)*1LL*(i-k)+max(n%i-k+,)-!k;
cout<<ans;
}

AtCoder Regular Contest 091的更多相关文章

  1. AtCoder Regular Contest 091&092

    091E(构造) 题意: 给出n,a,b.你需要构造出一个长度为n的n的排列,其中最长上升子序列的长度为a,最长下降子序列的长度为b. n,a,,b<=3e5 分析: 我们可以构造出这样的数列, ...

  2. AtCoder Regular Contest 061

    AtCoder Regular Contest 061 C.Many Formulas 题意 给长度不超过\(10\)且由\(0\)到\(9\)数字组成的串S. 可以在两数字间放\(+\)号. 求所有 ...

  3. AtCoder Regular Contest 094 (ARC094) CDE题解

    原文链接http://www.cnblogs.com/zhouzhendong/p/8735114.html $AtCoder\ Regular\ Contest\ 094(ARC094)\ CDE$ ...

  4. AtCoder Regular Contest 092

    AtCoder Regular Contest 092 C - 2D Plane 2N Points 题意: 二维平面上给了\(2N\)个点,其中\(N\)个是\(A\)类点,\(N\)个是\(B\) ...

  5. AtCoder Regular Contest 093

    AtCoder Regular Contest 093 C - Traveling Plan 题意: 给定n个点,求出删去i号点时,按顺序从起点到一号点走到n号点最后回到起点所走的路程是多少. \(n ...

  6. AtCoder Regular Contest 094

    AtCoder Regular Contest 094 C - Same Integers 题意: 给定\(a,b,c\)三个数,可以进行两个操作:1.把一个数+2:2.把任意两个数+1.求最少需要几 ...

  7. AtCoder Regular Contest 095

    AtCoder Regular Contest 095 C - Many Medians 题意: 给出n个数,求出去掉第i个数之后所有数的中位数,保证n是偶数. \(n\le 200000\) 分析: ...

  8. AtCoder Regular Contest 102

    AtCoder Regular Contest 102 C - Triangular Relationship 题意: 给出n,k求有多少个不大于n的三元组,使其中两两数字的和都是k的倍数,数字可以重 ...

  9. AtCoder Regular Contest 096

    AtCoder Regular Contest 096 C - Many Medians 题意: 有A,B两种匹萨和三种购买方案,买一个A,买一个B,买半个A和半个B,花费分别为a,b,c. 求买X个 ...

随机推荐

  1. JavaScript命名——name不能做变量名

    使用name作为变量名(var name = ‘’),在IE中未引起bug,在Chrome中引起bug但未明确指出命名错误,而是会报其他错误,故不便于发现. 现象原因: javascript中name ...

  2. 【TensorFlow入门完全指南】神经网络篇·卷积神经网络

    加载数据集. 这里的keep_prob是dropout的一个参数.dropout是一种随机置零的策略,用来防止模型过拟合. 这里定义两层,上面是卷积层,下面是池化层. 搭建了一层卷积.一层池化.一层卷 ...

  3. 2018.5.4 AndroidStudio遇到的问题

    新建项目初出现异常报错 Error:Execution failed for task ':app:preDebugAndroidTestBuild'. > Conflict 发生这类型的错误, ...

  4. J.U.C知识点梳理

    java.util.concurrent : 提供了并发编程的解决方案 1.CAS 是java.util.concurrent.atomic包的基础 2.AQS是java.util.concurren ...

  5. bower使用

    1,先安装nodejs(npm),Git 2,安装bower cmd执行到在项目文件夹下路径,执行npm install bower 3,执行bower init  项目根目录下将生成bower.js ...

  6. mysql5.7.24 解压版安装步骤以及遇到的问题

    1.下载 https://dev.mysql.com/downloads/mysql/ 2.解压到固定位置,如D:\MySQL\mysql-5.7.24 3.添加my.ini文件 跟bin同级 [my ...

  7. Java基础面试操作题:线程同步代码块 两个客户往一个银行存钱,每人存三十次一次存一百。 模拟银行存钱功能,时时银行现金数。

    package com.swift; public class Bank_Customer_Test { public static void main(String[] args) { /* * 两 ...

  8. 【数学 BSGS】bzoj2242: [SDOI2011]计算器

    数论的板子集合…… Description 你被要求设计一个计算器完成以下三项任务: 1.给定y,z,p,计算Y^Z Mod P 的值: 2.给定y,z,p,计算满足xy≡ Z ( mod P )的最 ...

  9. destoon 给超级管理员系统权限(管理员管理,日志管理等)

    destoon 后台某些系统权限除了网站创始人之外其他超管事没有权限的,现需要给其他超级管理员添加普通管理员的权限. 1.首先  admin/global.func.php  admin_check函 ...

  10. url,href,src区别

    URL(Uniform Resource Locator) 统一资源定位符是对可以从互联网上得到的资源的位置和访问方法的一种简洁的表示,是互联网上标准资源的地址.互联网上的每个文件都有一个唯一的URL ...