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 ...
- JavaScript事件模拟元素拖动
一.前言: 最近要实现一个元素拖放效果,鼠标拖动元素并且定位元素,首先想到的是HTML5中的拖放,在HTML5中,有一个draggable属性,且有dragstart, dragover, drop等 ...
- javascript私有静态成员
就私有静态成员而言,指的是成员具有如下属性:1.以同一个构造函数创建的所有对象共享该成员.2.构造函数外部不可访问该成员. //构造函数 var Gadget = (function(){ //静态变 ...
- JavaScript继承详解(三)
在第一章中,我们使用构造函数和原型的方式在JavaScript的世界中实现了类和继承, 但是存在很多问题.这一章我们将会逐一分析这些问题,并给出解决方案. 注:本章中的jClass的实现参考了Simp ...
- J2EE简介
一,J2EE概念: J2EE的全称为,Java2 Platform Enterprise Edition,Java或java2平台企业版,他是基于java平台或java2平台的标准版,保留并扩展了J2 ...
- 20155206 2016-2017-2 《Java程序设计》第7周学习总结
20155206 2016-2017-2 <Java程序设计>第7周学习总结 教材学习内容总结 认识时间与日期 1.格林威治时间(GMT):通过观察太阳而得,因为地球公转轨道为椭圆形且速度 ...
- 洛谷 P3916 【图的遍历】反向加边+dfs
前言: 对于这类带环的图,一般记忆化搜索不能很好的对所有遍历的边进行更新取值.因为环上的点可以相互到达,所以他们的答案因当是同步更新的,而dfs一旦你回溯完环上某个点就不会在更新这个点的答案了,做不到 ...
- Redis持久化——RDB快照
一.是什么? 在指定的时间间隔内将内存中的数据集快照写入磁盘,也就是行话讲的Snapshot快照,它恢复时是将快照文件直接读到内存里. Redis会单独创建(fork)一个子进程来进行持久化,会先将数 ...
- Linux内存管理 【转】
转自:http://blog.chinaunix.net/uid-25909619-id-4491368.html Linux内存管理 摘要:本章首先以应用程序开发者的角度审视Linux的进程内存管理 ...
- linux下查看各硬件型号
查看主板型号 # dmidecode |grep -A 8 "System Information"System Information 上网查DELL CS24-TY,找到说主板 ...