UESTCOJ-BiliBili, ACFun… And More!(水题)
BiliBili, ACFun… And More!
Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)
Some of you may have already noticed, there is a team in our final contest whose name is UESTC_BiliBilii, with user id as ACfun. Actually, both of them are websites mainly for watching videos.
So in this problem we also deal with video-share websites. When watching videos online, two numbers are very important. One is the playing speed: the speed you play the video, say X KB per second. The other is the downloading speed: the speed the computer downloads the video from the internet, say Y KB per second. Obviously, if X>Y, then you may have to pause for some time, since you cannot play something that hasn’t been downloaded!
The playing speed can also be described as the moving speed of the circle at the bottom of the videos, see the pictures below.

The circle will move along the blue bar, which is full now, indicating that downloading is already complete.
In this problem, we suppose that X and Y will always be constant.
Kennethsnow has a special habit when watching videos, let me tell you. First of all, he will wait for some time to download part of the video, say T seconds. Then, he starts to play the video.
If at a certain time, the video is paused, then kennethsnow will move the cursor(The circle in the picture) instantly to the leftmost position! That means, he will watch the video again, from the very beginning.
He will do this again and again, until the video comes to an end. Given X, Y, T, and the total size of the video, what is the time kennethsnow needs, to finish his watching?
Input
The first line of input contains a number T, indicating the number of test cases. (T≤1000).For each case, there will be four integers X, Y, T and S, which is the playing speed, downloading speed, the time kennethsnow will wait before playing, and the total size of video, given in KB. (1≤X,Y,T≤20, 1≤S≤100).
Output
For each case, output Case #i: first. (i is the number of the test case, from 1 to T). Then output the time kennethsnow needs to finish watching, in decimals, round to 3 decimal places.
Sample input and output
| Sample Input | Sample Output |
|---|---|
3 |
Case #1: 10.000 |
好久没有刷题了,最近又因为考试,就只能偶尔刷刷水题。题意不说了,在纸上画一下利用物理知识就能解决了。
#include<stdio.h>
int main ()
{
int i=;
double t0,x,y,t,s;
double time,t1,xTime,yTime,catchtime;
scanf("%lf",&t0);
while(t0--){
scanf("%lf%lf%lf%lf",&x,&y,&t,&s);
time=;
xTime=;
yTime=t;
if(x<=y) time+=s/x;
else{
catchtime=y*t/(x-y);
while(catchtime<s/x){
xTime+=catchtime;
yTime+=catchtime;
catchtime=y*yTime/(x-y);
}
time+=xTime+s/x;
}
printf("Case #%d: %.3f\n",i++,time);
}
return ;
}
UESTCOJ-BiliBili, ACFun… And More!(水题)的更多相关文章
- cdoj 03 BiliBili, ACFun… And More! 水题
Article Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/problem/show/3 Descr ...
- HDOJ 2317. Nasty Hacks 模拟水题
Nasty Hacks Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Tota ...
- ACM :漫漫上学路 -DP -水题
CSU 1772 漫漫上学路 Time Limit: 1000MS Memory Limit: 131072KB 64bit IO Format: %lld & %llu Submit ...
- ytu 1050:写一个函数,使给定的一个二维数组(3×3)转置,即行列互换(水题)
1050: 写一个函数,使给定的一个二维数组(3×3)转置,即行列互换 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 154 Solved: 112[ ...
- [poj2247] Humble Numbers (DP水题)
DP 水题 Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The se ...
- gdutcode 1195: 相信我这是水题 GDUT中有个风云人物pigofzhou,是冰点奇迹队的主代码手,
1195: 相信我这是水题 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 821 Solved: 219 Description GDUT中有个风云人 ...
- BZOJ 1303 CQOI2009 中位数图 水题
1303: [CQOI2009]中位数图 Time Limit: 1 Sec Memory Limit: 162 MBSubmit: 2340 Solved: 1464[Submit][Statu ...
- 第十一届“蓝狐网络杯”湖南省大学生计算机程序设计竞赛 B - 大还是小? 字符串水题
B - 大还是小? Time Limit:5000MS Memory Limit:65535KB 64bit IO Format: Description 输入两个实数,判断第一个数大 ...
- ACM水题
ACM小白...非常费劲儿的学习中,我觉得目前我能做出来的都可以划分在水题的范围中...不断做,不断总结,随时更新 POJ: 1004 Financial Management 求平均值 杭电OJ: ...
随机推荐
- Linux下压缩某个文件夹(文件夹打包)
tar -zcvf /home/xahot.tar.gz /xahottar -zcvf 打包后生成的文件名全路径 要打包的目录例子:把/xahot文件夹打包后生成一个/home/xahot.tar. ...
- Https 原理
HTTPS其实是有两部分组成:HTTP + SSL / TLS, 也就是在HTTP上又加了一层处理加密信息的模块.服务端和客户端的信息传输都会通过TLS进行加密,所以传输的数据都是加密后的数据 1. ...
- POJ_1631_Bridging_Signals_(动态规划,LIS)
描述 http://poj.org/problem?id=1631 铁路左右相连,要求去掉一些边,使得剩下的边不交叉,求剩余边数的最大值. Bridging signals Time Limit: 1 ...
- BZOJ2296: 【POJ Challenge】随机种子
2296: [POJ Challenge]随机种子 Time Limit: 1 Sec Memory Limit: 128 MBSec Special JudgeSubmit: 114 Solv ...
- Set up JBPM5.4 Final Installer to use MS SQL Server 2008 using JTDS(转)
[-] A What I Am Going To Do B The Setup Steps C Lets Install it A. What I Am Going To Do B. The Se ...
- ECSHOP首页调用指定分类下的商品
转:http://bbs.ecshop.com/thread-1123207-1-1.html 调用某个分类下的商品,方法有很多种的,不过都需要先在后台设置模板那里设置显示和显示条数, 然后在需要调用 ...
- Action中取得request,session的四种方式
Action中取得request,session的四种方式 在Struts2中,从Action中取得request,session的对象进行应用是开发中的必需步骤,那么如何从Action中取得这些对象 ...
- Design T-Shirt
Design T-Shirt Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
- leetcode 二分查找
https://oj.leetcode.com/problems/search-for-a-range/就是一个二分查找,没事练练手 public class Solution { public in ...
- JavaScript高级程序设计23.pdf
document对象作为HTMLDocument的一个实例,它还有一些标准的Document对象所没有的属性,这些属性提供了网页上的一些信息 //取得文档标题 var title1=document. ...