(code jam)Problem B. Cookie Clicker Alpha

本文介绍了一款名为CookieClicker的游戏,玩家通过点击巨型饼干获得饼干并购买建筑以增加饼干产量。文章详细分析了最优游戏策略,并提供了一个示例帮助理解如何在限定时间内获得指定数量的饼干。

Introduction

Cookie Clicker is a Javascript game by Orteil, where players click on a picture of a giant cookie. Clicking on the giant cookie gives them cookies. They can spend those cookies to buy buildings. Those buildings help them get even more cookies. Like this problem, the game is very cookie-focused. This problem has a similar idea, but it does not assume you have played Cookie Clicker. Please don't go play it now: it might be a long time before you come back.

Problem

In this problem, you start with 0 cookies. You gain cookies at a rate of 2 cookies per second, by clicking on a giant cookie. Any time you have at least C cookies, you can buy a cookie farm. Every time you buy a cookie farm, it costs you C cookies and gives you an extra F cookies per second.

Once you have X cookies that you haven't spent on farms, you win! Figure out how long it will take you to win if you use the best possible strategy.

Example

Suppose C=500.0, F=4.0 and X=2000.0. Here's how the best possible strategy plays out:

  1. You start with 0 cookies, but producing 2 cookies per second.
  2. After 250 seconds, you will have C=500 cookies and can buy a farm that producesF=4 cookies per second.
  3. After buying the farm, you have 0 cookies, and your total cookie production is 6 cookies per second.
  4. The next farm will cost 500 cookies, which you can buy after about 83.3333333seconds.
  5. After buying your second farm, you have 0 cookies, and your total cookie production is 10 cookies per second.
  6. Another farm will cost 500 cookies, which you can buy after 50 seconds.
  7. After buying your third farm, you have 0 cookies, and your total cookie production is 14 cookies per second.
  8. Another farm would cost 500 cookies, but it actually makes sense not to buy it: instead you can just wait until you have X=2000 cookies, which takes about142.8571429 seconds.
Total time: 250 + 83.3333333 + 50 + 142.8571429 = 526.1904762 seconds.

Notice that you get cookies continuously: so 0.1 seconds after the game starts you'll have 0.2 cookies, and π seconds after the game starts you'll have 2π cookies.

Input

The first line of the input gives the number of test cases, TT lines follow. Each line contains three space-separated real-valued numbers: CF and X, whose meanings are described earlier in the problem statement.

CF and X will each consist of at least 1 digit followed by 1 decimal point followed by from 1 to 5 digits. There will be no leading zeroes.

Output

For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1) and y is the minimum number of seconds it takes before you can have X delicious cookies.

We recommend outputting y to 7 decimal places, but it is not required. y will be considered correct if it is close enough to the correct number: within an absolute or relative error of 10-6. See the FAQ for an explanation of what that means, and what formats of real numbers we accept.

Limits

1 ≤ T ≤ 100.

Small dataset

1 ≤ C ≤ 500.
1 ≤ F ≤ 4.
1 ≤ X ≤ 2000.

Large dataset

1 ≤ C ≤ 10000.
1 ≤ F ≤ 100.
1 ≤ X ≤ 100000.

Sample


Input 
 

Output 
 
4
30.0 1.0 2.0
30.0 2.0 100.0
30.50000 3.14159 1999.19990
500.0 4.0 2000.0

Case #1: 1.0000000
Case #2: 39.1666667
Case #3: 63.9680013
Case #4: 526.1904762

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
using namespace std;
int main()
{
	int T,cas,i;
	double c,f,x;
	freopen("outpu.txt","w",stdout);
	scanf("%d",&T);
	for (cas=1;cas<=T;cas++)
	{
		scanf("%lf%lf%lf",&c,&f,&x);
		double ans=x/2.0;
		double k=2.0;
		double s=0;
		for (i=1;i<=100000;i++)
		{
			s+=c/k;
			k+=f;
			double q=x/k+s;
		//	cout<<q<<endl;
			if (q>ans) break;
			else ans=q;
		}
		printf("Case #%d: %.7lf\n",cas,ans);
	}
	return 0;
}



内容概要:本文档介绍了基于3D FDTD(时域有限差分)方法在MATLAB平台上对微带线馈电的矩形天线进行仿真分析的技术方案,重点在于模拟超MATLAB基于3D FDTD的微带线馈矩形天线分析[用于模拟超宽带脉冲通过线馈矩形天线的传播,以计算微带结构的回波损耗参数]宽带脉冲信号通过天线结构的传播过程,并计算微带结构的回波损耗参数(S11),以评估天线的匹配性能和辐射特性。该方法通过建立三维电磁场模型,精确求解麦克斯韦方程组,适用于高频电磁仿真,能够有效分析天线在宽频带内的响应特性。文档还提及该资源属于一个涵盖多个科研方向的综合性MATLAB仿真资源包,涉及通信、信号处理、电力系统、机器学习等多个领域。; 适合人群:具备电磁场与微波技术基础知识,熟悉MATLAB编程及数值仿真的高校研究生、科研人员及通信工程领域技术人员。; 使用场景及目标:① 掌握3D FDTD方法在天线仿真中的具体实现流程;② 分析微带天线的回波损耗特性,优化天线设计参数以提升宽带匹配性能;③ 学习复杂电磁问题的数值建模与仿真技巧,拓展在射频与无线通信领域的研究能力。; 阅读建议:建议读者结合电磁理论基础,仔细理解FDTD算法的离散化过程和边界条件设置,运行并调试提供的MATLAB代码,通过调整天线几何尺寸和材料参数观察回波损耗曲线的变化,从而深入掌握仿真原理与工程应用方法。
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