D. Dreamoon and Sets
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Dreamoon likes to play with sets, integers and  is
defined as the largest positive integer that divides both a and b.

Let S be a set of exactly four distinct integers greater than 0.
Define S to be of rank k if and only if for all
pairs of distinct elements sisj fromS.

Given k and n, Dreamoon wants to make up n sets
of rank k using integers from 1 to m such
that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print
one possible solution.

Input

The single line of the input contains two space separated integers nk (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).

Output

On the first line print a single integer — the minimal possible m.

On each of the next n lines print four space separated integers representing the i-th
set.

Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one
of them.

Sample test(s)
input
1 1
output
5
1 2 3 5
input
2 2
output
22
2 4 6 22
14 18 10 16
Note

For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since .

构造。当k等于1时,推几组数据。比如1,2,3,5;7,8,9,11;13,14,15,17。19,20,21,23;25,26,27,29。就会发现是以6为周期,而对每一个周期内的数乘以k就会使周期内的数两两的最大公约数为k。



代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int main()
{
int n, k;
scanf("%d %d", &n, &k);
int a = 1, b = 2, c = 3, d = 5;
printf("%d\n", (d * k + 6 * k * (n- 1)));
a*=k;
b*=k;
c*=k;
d*=k;
for(int i = 0; i < n; i++)
{
printf("%d %d %d %d\n",a, b, c, d);
a += 6 * k;
b += 6 * k;
c += 6 * k;
d += 6 * k;
}
}



D. Dreamoon and Sets(Codeforces Round #272)的更多相关文章

  1. B. Dreamoon and WiFi(Codeforces Round 272)

    B. Dreamoon and WiFi time limit per test 1 second memory limit per test 256 megabytes input standard ...

  2. A. Dreamoon and Stairs(Codeforces Round #272)

    A. Dreamoon and Stairs time limit per test 1 second memory limit per test 256 megabytes input standa ...

  3. E. Dreamoon and Strings(Codeforces Round #272)

    E. Dreamoon and Strings time limit per test 1 second memory limit per test 256 megabytes input stand ...

  4. Codeforces Round #272 (Div. 2) 题解

    Codeforces Round #272 (Div. 2) A. Dreamoon and Stairs time limit per test 1 second memory limit per ...

  5. Codeforces Round #272 (Div. 2) D. Dreamoon and Sets 构造

    D. Dreamoon and Sets 题目连接: http://www.codeforces.com/contest/476/problem/D Description Dreamoon like ...

  6. Codeforces Round #272 (Div. 2) D.Dreamoon and Sets 找规律

    D. Dreamoon and Sets   Dreamoon likes to play with sets, integers and .  is defined as the largest p ...

  7. Codeforces Round #272 (Div. 2)AK报告

    A. Dreamoon and Stairs time limit per test 1 second memory limit per test 256 megabytes input standa ...

  8. 「日常训练」Watering Flowers(Codeforces Round #340 Div.2 C)

    题意与分析 (CodeForces 617C) 题意是这样的:一个花圃中有若干花和两个喷泉,你可以调节水的压力使得两个喷泉各自分别以\(r_1\)和\(r_2\)为最远距离向外喷水.你需要调整\(r_ ...

  9. 「日常训练」Alternative Thinking(Codeforces Round #334 Div.2 C)

    题意与分析 (CodeForces - 603A) 这题真的做的我头疼的不得了,各种构造样例去分析性质... 题意是这样的:给出01字符串.可以在这个字符串中选择一个起点和一个终点使得这个连续区间内所 ...

随机推荐

  1. [HAOI2006]旅行(并查集)

    寒假填坑五十道省选题——第五道 [HAOI2006]旅行 题目描述 Z小镇是一个景色宜人的地方,吸引来自各地的观光客来此旅游观光.Z小镇附近共有N个景点(编号为1,2,3,…,N),这些景点被M条道路 ...

  2. PKU 2411 Mondriaan's Dream 状态DP

    以前做过这题,今天又写了一次,突然发现写了一个好漂亮的DFS……(是不是太自恋了 - -#) 代码: #include <cstdio> #include <cstring> ...

  3. 【Round #36 (Div. 2 only) C】Socks Pairs

    [题目链接]:https://csacademy.com/contest/round-36/task/socks-pairs/ [题意] 给你n种颜色的袜子,每种颜色颜色的袜子有ai只; 假设你在取袜 ...

  4. 【cdoj 1544】当咸鱼也要按照基本法

    [题目链接]:http://acm.uestc.edu.cn/#/problem/show/1544 [题意] [题解] 容斥原理题; 1..(2^m)-1枚举 设k为其二进制形式中1的个数; k为奇 ...

  5. 1193 Eason

    Eason Acceteped : 57   Submit : 129 Time Limit : 1000 MS   Memory Limit : 65536 KB Description 题目描述 ...

  6. ArcGIS api for javascript——查询,立刻打开信息窗口

    描述 本例展示了当一个要素被查询时如何立刻打开一个InfoWindow.信息窗口能被用来将要素的属性格式化成用户易读的格式. 本例中,地图和查询任务都使用ESRI sample server上的服务K ...

  7. Android 使用DrawerLayout高速实现側滑菜单

    一.概述 DrawerLayout是一个能够方便的实现Android側滑菜单的组件,我近期开发的项目中也有一个側滑菜单的功能.于是DrawerLayout就派上用场了.假设你从未使用过DrawerLa ...

  8. [Angular] *ngIf syntx

    <div class="profile"> <img [src]="user.img" width="50px"> ...

  9. iOS学习9_事件分发&amp;响应链

    iOS的三种事件:触摸事件/运动事件/远程控制事件 typedef enum { UIEventTypeTouches, UIEventTypeMotion, UIEventTypeRemoteCon ...

  10. Linux LVM(逻辑卷管理)

    Lvm基本应用 什么是LVM? LVM 的全称是 Logical Volume Manager.中文为逻辑卷管理.它是Linux对磁盘分区的一种管理机制.它在传统的硬盘(或硬盘分区)和文件系统之间建立 ...