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

Description

 

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?

Since World crashes too often, now he is asking his friend ATM for the optimal strategy to input his article. A strategy is measured by its expectation keys DRD needs to press.

Note that DRD can press a key at fast enough speed

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

3 1 1 2 10 2 1 3 20 3 1 4 30

Sample Output

Case #1: 10.000

Case #2: 19.000

Case #3: 26.250

HINT

题意

有一个人在bilibili看动漫，看的速度是x，下载的速度是y，一开始等了t秒再播放的，视频总大小为x

然后这个人有个怪癖，就是当出现卡顿的时候，他就会把进度条重新拉回开始位置

然后问你得过多少秒，才能看完整个视频

题解：

高中物理题，随便推推就好啦～

代码：

 

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define test freopen("test.txt","r",stdin)  

#define maxn 200

#define mod 10007

#define eps 1e-9

int Num;

char CH[20];

const int inf=0x3f3f3f3f;

const ll infll = 0x3f3f3f3f3f3f3f3fLL;

inline ll read()

{

    ll x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

inline void P(int x)

{

    Num=0;if(!x){putchar('0');puts("");return;}

    while(x>0)CH[++Num]=x%10,x/=10;

    while(Num)putchar(CH[Num--]+48);

    puts("");

}

//**************************************************************************************



double x,y,t,z;

double cal(double v1,double v2,double s)

{

    return s/(v1-v2);

}

void solve()

{

    cin>>x>>y>>t>>z;

    if(y>=x)

    {

        printf("%.3lf\n",z/x);

        return;

    }

    double ans=0;

    double kiss=y*t;

    while(cal(x,y,kiss)<z/x)

    {

        ans+=cal(x,y,kiss);

        kiss=cal(x,y,kiss)*x;

    }

    printf("%.3lf\n",ans+z/x);

}

int main()

{

    //test;

    int t=read();

    for(int cas=1;cas<=t;cas++)

    {

        printf("Case #%d: ",cas);

        solve();

    }

}