Codeforces Round #372 (Div. 1) A. Plus and Square Root 数学题
A. Plus and Square Root
题目连接:
http://codeforces.com/contest/715/problem/A
Description
ZS the Coder is playing a game. There is a number displayed on the screen and there are two buttons, ' + ' (plus) and '' (square root). Initially, the number 2 is displayed on the screen. There are n + 1 levels in the game and ZS the Coder start at the level 1.
When ZS the Coder is at level k, he can :
Press the ' + ' button. This increases the number on the screen by exactly k. So, if the number on the screen was x, it becomes x + k.
Press the '' button. Let the number on the screen be x. After pressing this button, the number becomes . After that, ZS the Coder levels up, so his current level becomes k + 1. This button can only be pressed when x is a perfect square, i.e. x = m2 for some positive integer m.
Additionally, after each move, if ZS the Coder is at level k, and the number on the screen is m, then m must be a multiple of k. Note that this condition is only checked after performing the press. For example, if ZS the Coder is at level 4 and current number is 100, he presses the '' button and the number turns into 10. Note that at this moment, 10 is not divisible by 4, but this press is still valid, because after it, ZS the Coder is at level 5, and 10 is divisible by 5.
ZS the Coder needs your help in beating the game — he wants to reach level n + 1. In other words, he needs to press the '' button n times. Help him determine the number of times he should press the ' + ' button before pressing the '' button at each level.
Please note that ZS the Coder wants to find just any sequence of presses allowing him to reach level n + 1, but not necessarily a sequence minimizing the number of presses.
Input
The first and only line of the input contains a single integer n (1 ≤ n ≤ 100 000), denoting that ZS the Coder wants to reach level n + 1.
Output
Print n non-negative integers, one per line. i-th of them should be equal to the number of times that ZS the Coder needs to press the ' + ' button before pressing the '' button at level i.
Each number in the output should not exceed 1018. However, the number on the screen can be greater than 1018.
It is guaranteed that at least one solution exists. If there are multiple solutions, print any of them.
Sample Input
4
Sample Output
2
17
46
97
Hint
题意
一开始屏幕上是x,然后你有两个操作
操作一是让数字加上(x-1)
操作二是让数字开根号,但是数字开根号后,必须是x的倍数
现在假设一开始屏幕上数字是2,你想让数字变成n+1,问你每一步你需要进行操作一多少次。
题解:
数学题,每次把最小的那个满足的数字拿出来,就会发现有规律,然后输出就好了。
代码
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
scanf("%d",&n);
cout<<"2"<<endl;
for(int i=2;i<=n;i++)
cout<<1ll*i*(i+1)*(i+1)-i+1<<endl;
}
Codeforces Round #372 (Div. 1) A. Plus and Square Root 数学题的更多相关文章
- Codeforces Round #372 (Div. 2) C. Plus and Square Root
题目链接 分析:这题都过了2000了,应该很简单..写这篇只是为了凑篇数= = 假设在第级的时候开方过后的数为,是第级的系数.那么 - 显然,最小的情况应该就是, 化简一下公式,在的情况下应该是,注意 ...
- Codeforces Round #372 (Div. 2)
Codeforces Round #372 (Div. 2) C. Plus and Square Root 题意 一个游戏中,有一个数字\(x\),当前游戏等级为\(k\),有两种操作: '+'按钮 ...
- Codeforces Round #372 (Div. 2) A .Crazy Computer/B. Complete the Word
Codeforces Round #372 (Div. 2) 不知不觉自己怎么变的这么水了,几百年前做A.B的水平,现在依旧停留在A.B水平.甚至B题还不会做.难道是带着一种功利性的态度患得患失?总共 ...
- 构造水题 Codeforces Round #206 (Div. 2) A. Vasya and Digital Root
题目传送门 /* 构造水题:对于0的多个位数的NO,对于位数太大的在后面补0,在9×k的范围内的平均的原则 */ #include <cstdio> #include <algori ...
- Codeforces 715A & 716C Plus and Square Root【数学规律】 (Codeforces Round #372 (Div. 2))
C. Plus and Square Root time limit per test 2 seconds memory limit per test 256 megabytes input stan ...
- Codeforces 715B & 716D Complete The Graph 【最短路】 (Codeforces Round #372 (Div. 2))
B. Complete The Graph time limit per test 4 seconds memory limit per test 256 megabytes input standa ...
- Codeforces 716A Crazy Computer 【模拟】 (Codeforces Round #372 (Div. 2))
A. Crazy Computer time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...
- Codeforces 716B Complete the Word【模拟】 (Codeforces Round #372 (Div. 2))
B. Complete the Word time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- Codeforces Round #372 (Div. 2) C 数学
http://codeforces.com/contest/716/problem/C 题目大意:感觉这道题还是好懂得吧. 思路:不断的通过列式子的出来了.首先我们定义level=i, uplevel ...
随机推荐
- BSGS 算法
求解 A^x ≡ B mod C C是质数 的最小非负整数解 证明:A^x ≡ A^(x%φ(C)) mod C A^(x%φ(C)) ≡ A^(x-k*φ(C)) ≡ (A^x)/ A^(k*φ ...
- 第12月第1天 MASConstraintMaker crash
1. crash [valueLabel mas_makeConstraints:^(PAKitMASConstraintMaker *make) { make.left.equalTo(finish ...
- 利用iis虚拟目录实现文件服务器功能
要求说明: 通过网站上传文件保存到统一的文件服务器上. 服务器说明: 1.文件服务器以下称为FilesServer,IP地址为:192.168.1.213 2.Web服务器为以下称为WebServer ...
- Linux系统的快速启动机制(内核切换) 【转】
转自:http://blog.chinaunix.net/xmlrpc.php?r=blog/article&uid=26807463&id=4187846 原文地址:Linux系统的 ...
- WPF的EventAggregator的发布和订阅
EventAggregator是Prism中专门处理ViewModel与ViewModel之间事件传递的类对象,它提供了针对事件的发布方法和订阅方法,所以可以非常方便的来管理事件.下面分几步来实现相关 ...
- 002_CentOS-6.4-x86_64安装包的说明
http://mirrors.sohu.com/centos/6.6/isos/x86_64/?qq-pf-to=pcqq.group //souhu镜像下载地址 0_README.txt 25-Oc ...
- Flask页面模板化代码片段
文中展示了Flask官网的Tutorial示例中用到的模板化代码片段,以备查阅. base.html: {% block title %}{% endblock %} {% if g.user %} ...
- win7(x64)安装scrapy框架
Scrapy(官网http://scrapy.org/)是Python开发网络爬虫,一个极好的开源工具.本次安装Scrapy确实不易啊.居然花了2天多时间,需要的支持包比较多,这些支持包相互之间的依赖 ...
- String,StringBuffer和StringBuilder的区别
(1)String类的API概述是这样的:String类代表字符串,Java程序中的所有字符串字面值都作为此类的实例体现.字符串是常量,它们的值在创建之后不能更改.可见,String是对象且为不可变对 ...
- pycharm+PyQt5+python最新开发环境配置
Python 3.6https://www.python.org/downloads/windows/========================================PyQt5 pip ...