D - Decrease (Contestant ver.)


Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N−1 or smaller.

  • Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.

It can be proved that the largest element in the sequence becomes N−1 or smaller after a finite number of operations.

You are given an integer K. Find an integer sequence ai such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.

Constraints

  • 0≤K≤50×1016

Input

Input is given from Standard Input in the following format:

K

Output

Print a solution in the following format:

N
a1 a2 ... aN

Here, 2≤N≤50 and 0≤ai≤1016+1000 must hold.


Sample Input 1

Copy
0

Sample Output 1

Copy
4
3 3 3 3

Sample Input 2

Copy
1

Sample Output 2

Copy
3
1 0 3

Sample Input 3

Copy
2

Sample Output 3

Copy
2
2 2

The operation will be performed twice: [2, 2] -> [0, 3] -> [1, 1].


Sample Input 4

Copy
3

Sample Output 4

Copy
7
27 0 0 0 0 0 0

Sample Input 5

Copy
1234567894848

Sample Output 5

Copy
10
1000 193 256 777 0 1 1192 1234567891011 48 425

直接定义数组长为50,k=0:0,1,2,............49;
k=1:50,0,1,2,3,..............48;
k=2:49,50,1,2,3..............47;
.......
k=50:1,2,3,..............48,49,50;
周期为50
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll n;
ll a[];
int main()
{
scanf("%lld",&n);
for(int i=;i<=;i++)
{
a[i]=n/+i-;
}
n%=;
for(int i=;i<=n;i++)
{
for(int j=;j<=;j++)
{
a[j]--;
}
a[i]+=;
}
printf("50\n");
for(int i=;i<=;i++)
{
if(i!=) printf(" ");
printf("%lld",a[i]);
}
printf("\n");
return ;
}

Atcoder At Beginner Contest 068 D - Decrease (Contestant ver.)的更多相关文章

  1. 【构造】AtCoder Regular Contest 079 D - Decrease (Contestant ver.)

    从n个t变化到n个t-1,恰好要n步,并且其中每一步的max值都>=t,所以把50个49当成最终局面,从这里开始,根据输入的K计算初始局面即可. #include<cstdio> # ...

  2. Atcoder At Beginner Contest 068 C - Cat Snuke and a Voyage

    C - Cat Snuke and a Voyage Time limit : 2sec / Memory limit : 256MB Score : 300 points Problem State ...

  3. AtCoder Beginner Contest 068 ABCD题

    A - ABCxxx Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement This contes ...

  4. AtCoder Beginner Contest 068

    A - ABCxxx 题意: 给出n,输出“ABCn”就可以了,纯水题. B - Break Number 题意: 给出n,找出从1到n的闭区间内能够被2整除最多次的数. 思路: 直接模拟. 代码: ...

  5. Atcoder arc079 D Decrease (Contestant ver.) (逆推)

    D - Decrease (Contestant ver.) Time limit : 2sec / Memory limit : 256MB Score : 600 points Problem S ...

  6. Atcoder AtCoder Regular Contest 079 E - Decrease (Judge ver.)

    E - Decrease (Judge ver.) Time limit : 2sec / Memory limit : 256MB Score : 600 points Problem Statem ...

  7. 【贪心】AtCoder Regular Contest 079 E - Decrease (Judge ver.)

    每次将最大的数减到n以下,如此循环直到符合题意. 复杂度大概是n*n*log?(?). #include<cstdio> #include<iostream> #include ...

  8. atcoder beginner contest 251(D-E)

    Tasks - Panasonic Programming Contest 2022(AtCoder Beginner Contest 251)\ D - At Most 3 (Contestant ...

  9. AtCoder Beginner Contest 100 2018/06/16

    A - Happy Birthday! Time limit : 2sec / Memory limit : 1000MB Score: 100 points Problem Statement E8 ...

随机推荐

  1. iOS开发系列之四 - UITextView 使用方法小结

    // 初始化输入框并设置位置和大小 UITextView *textView = [[UITextView alloc] initWithFrame:CGRectMake(10, 10, 300, 1 ...

  2. PIO Core

    PIO核概述 具有Avalon接口的并行输入/输出(parallel input/output - PIO)核,在Avalon存储器映射(Avalon Memory-Mapped Avalon-MM) ...

  3. 4.angularJS-指令(directive)

    转自:https://www.cnblogs.com/best/p/6225621.html 指令(directive)是AngularJS模板标记和用于支持的JavaScript代码的组合.Angu ...

  4. 99.重载[] * -> ->*

    #include "mainwindow.h" #include <QApplication> #include <QPushButton>> //重 ...

  5. 基于Linux下Iptables限制BT下载的研究

    基于Linux下Iptables限制BT下载的研究   摘要:     当前BT下载技术和软件飞速发展,给人们网上冲浪获取资源带来了极大的便利, 但同时BT占用大量的网络带宽等资源也给网络和网络管理员 ...

  6. selenium 窗口句柄之间的切换

    以前使用selenium时都是在单窗口的模式下,本次新增多窗口下的窗口之间切换 from selenium import webdriver from selenium.webdriver.commo ...

  7. git pull 、git fetch、 git clone

    git clone 代表从远程克隆过来包括所有的版本信息 git fetch是从远程获取最新的版本 git pull相当于 git fetch 然后再git merge

  8. 学习推荐《从Excel到Python数据分析进阶指南》高清中文版PDF

    Excel是数据分析中最常用的工具,本书通过Python与Excel的功能对比介绍如何使用Python通过函数式编程完成Excel中的数据处理及分析工作.在Python中pandas库用于数据处理,我 ...

  9. 安装php-ldap模块

    php-ldap模块作用就是实现ldap认证,因此需要安装 1.安装软件包解决依赖 yum install openldapyum install openldap-devel 2.拷贝库文件 cp ...

  10. jQuery获取区间随机数

    1.自定义函数 function getRandom(min,max){    //x上限,y下限    var x = max;    var y = min;    if(x<y){     ...