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. db2数据库备份

    一.离线备份 db2  list  database  directory -----查看有哪些数据库,确定需要备份哪个数据库 db2  disconnect  current -----断开以数据库 ...

  2. BZOJ 2539: [Ctsc2000]丘比特的烦恼

    Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 695  Solved: 260[Submit][Status][Discuss] Description ...

  3. ABAP的Package interface, 安卓的manifest.xml和Kubernetes的Capabilities

    ABAP 事务码SE21创建ABAP包接口.这是ABAP基于包层面的访问控制实现逻辑.包里可以存储很多ABAP对象.如果开发人员想将某些对象声明为包外程序也能访问,可以将这些对象放在包接口的Visib ...

  4. mdns小结

    mdns的功能和普通DNS很类似,即提供主机名到IP地址的解析服务.   mdns一些基本特性: 1,mdns主要为小型私有网络(不存在DNS)提供名称解析. 2,mdns使用多播(Multicast ...

  5. uva10129 PlayOnWords(并查集,欧拉回路)

    判断无向图是否存在欧拉回路,就是看度数为奇数的点有多少个,如果有两个,那么以那分别两个点为起点和终点,可以构造出一条欧拉回路,如果没有,就任意来,否则,欧拉回路不存在. 首先用并查集判断连通,然后统计 ...

  6. UVA 10570 Meeting with Aliens 外星人聚会

    题意:给你一个排列,每次可以交换两个整数(不一定要相邻),求最少交换次数把排列变成一个1~n的环形排列.(正反都算) 其实就是找环了,对于一个链状序列,最小交换次数等于不在对应位置的数字个数减去环的个 ...

  7. [学习笔记] SSD代码笔记 + EifficientNet backbone 练习

    SSD代码笔记 + EifficientNet backbone 练习 ssd代码完全ok了,然后用最近性能和速度都非常牛的Eifficient Net做backbone设计了自己的TinySSD网络 ...

  8. 小技巧:unicode RLO

    unicode 控制字符 RLO 可以将位于其后的文字翻转. 于是可以被病毒利用. 如图 重命名文件,在gpj前插入unicode RLO,之后若不小心,可能会被欺骗,误以为是jpg文件. 如果修改程 ...

  9. opensue "Have a lot of fun..."的出处

    每次登陆opensuse都会出现“Have a lot of fun...”,觉得奇怪. 通过搜索发现在这是/etc/motd文件中配置的. MOTD(5)                       ...

  10. 关于flyme5显示不到和卸载不到旧应用解决方法

    笔者买入一台mx5,升级flyme5后旧应用没有显示出来,而且在设置的应用管理都没显示旧应用. 通过adb命令: adb shell pm list packages显示所有包名, 查看自己要删除应用 ...