You're trying to set the record on your favorite video game. The game consists of N levels, which must be completed sequentially in order to beat the game. You usually complete each level as fast as possible, but sometimes finish a level slower. Specifically, you will complete the i-th level in either Fi seconds or Si seconds, where Fi < Si, and there's a Pi percent chance of completing it in Fi seconds. After completing a level, you may decide to either continue the game and play the next level, or reset the game and start again from the first level. Both the decision and the action are instant.

Your goal is to complete all the levels sequentially in at most R total seconds. You want to minimize the expected amount of time playing before achieving that goal. If you continue and reset optimally, how much total time can you expect to spend playing?

Input

The first line of input contains integers N and R , the number of levels and number of seconds you want to complete the game in, respectively. N lines follow. The ith such line contains integers Fi, Si, Pi (1 ≤ Fi < Si ≤ 100, 80 ≤ Pi ≤ 99), the fast time for level i, the slow time for level i, and the probability (as a percentage) of completing level i with the fast time.

Output

Print the total expected time. Your answer must be correct within an absolute or relative error of 10 - 9.

Formally, let your answer be a, and the jury's answer be b. Your answer will be considered correct, if .

Examples
input
1 82 8 81
output
3.14
input
2 3020 30 803 9 85
output
31.4
input
4 31963 79 8979 97 9175 87 8875 90 83
output
314.159265358
Note

In the first example, you never need to reset. There's an 81% chance of completing the level in 2 seconds and a 19% chance of needing 8 seconds, both of which are within the goal time. The expected time is 0.81·2 + 0.19·8 = 3.14.

In the second example, you should reset after the first level if you complete it slowly. On average it will take 0.25 slow attempts before your first fast attempt. Then it doesn't matter whether you complete the second level fast or slow. The expected time is 0.25·30 + 20 + 0.85·3 + 0.15·9 = 31.4.


  题目大意 一个人打游戏,需要不超过$R$秒通过$n$关,第$i$关有$P_{i}$的概率用$F_{i}$秒通过,$\left(1 - P_{i}\right)$的概率用$S_{i}$通过($F_{i} < S_{i}$),通过每一关可以选择重置游戏,然后从头开始,或者去打下一关。问不超过$R$秒通过所有关卡的期望耗时。

  转移是显然的。(如果这个都不会,请自定百度“概率dp入门题”)

  然后发现转移有环,还要做决策?

  然后列方程吧。。开心地发现不会解。

  可惜这里是信息学竞赛,不是数学竞赛。由于转移都需要 dp[][] 但是开始不知道它,所以考虑二分它,然后和推出来的 dp[][] 作比较。

  经过各种瞎猜和乱搞,可以发现一个神奇的事情

  然后就可根据它来确定一次check后,二分的范围。

  另外,由于坑人的精度问题,所以最好不要写while (l + eps < r) ,总之我这么写各种因为精度问题的TLE来了。

Code

 /**
  * Codeforces
  * Problem#866C
  * Accepted
  * Time: 62ms
  * Memory: 4316k
  */
 #include <bits/stdc++.h>
 using namespace std;
 typedef bool boolean;

 ;
 ;

 int n, R;
 int *fs, *ss;
 double *ps;

 inline void init() {
     scanf("%d%d", &n, &R);
     fs = )];
     ss = )];
     ps = )];
     ; i <= n; i++) {
         scanf("%d%d", fs + i, ss + i);
         cin >> ps[i];
         ps[i] *= 0.01;
     }
 }

 boolean vis[][];
 ][];

 double dfs(int d, int t, double &mid) {
     );
     if(vis[d][t])    return f[d][t];
     vis[d][t] = true;
     f[d][t] = (dfs(d + , t + fs[d + ], mid) + fs[d + ]) * ps[d + ] + (dfs(d + , t + ss[d + ], mid) + ss[d + ]) * ( - ps[d + ]);
     if(mid < f[d][t])    f[d][t] = mid;
     return f[d][t];
 }

 double dp(double mid) {
     memset(vis, false, sizeof(vis));
     , , mid);
 }

 inline void solve() {
     , r = 1e9;
     ; i < binary_lim; i++) {
         ;
         if(dp(mid) < mid)    r = mid;
         else    l = mid;
     }
     printf("%.9lf", l);
 }

 int main() {
     init();
     solve();
     ;
 }

Codeforces 866C Gotta Go Fast - 动态规划 - 概率与期望 - 二分答案的更多相关文章

  1. [Codeforces 865C]Gotta Go Fast(期望dp+二分答案)

    [Codeforces 865C]Gotta Go Fast(期望dp+二分答案) 题面 一个游戏一共有n个关卡,对于第i关,用a[i]时间通过的概率为p[i],用b[i]通过的时间为1-p[i],每 ...

  2. Codeforces 865C Gotta Go Fast 二分 + 期望dp (看题解)

    第一次看到这种骚东西, 期望还能二分的啊??? 因为存在重置的操作, 所以我们再dp的过程中有环存在. 为了消除环的影响, 我们二分dp[ 0 ][ 0 ]的值, 与通过dp得出的dp[ 0 ][ 0 ...

  3. bzoj 4318 OSU! - 动态规划 - 概率与期望

    Description osu 是一款群众喜闻乐见的休闲软件.  我们可以把osu的规则简化与改编成以下的样子:  一共有n次操作,每次操作只有成功与失败之分,成功对应1,失败对应0,n次操作对应为1 ...

  4. bzoj 4008 亚瑟王 - 动态规划 - 概率与期望

    Description 小 K 不慎被 LL 邪教洗脑了,洗脑程度深到他甚至想要从亚瑟王邪教中脱坑. 他决定,在脱坑之前,最后再来打一盘亚瑟王.既然是最后一战,就一定要打得漂 亮.众所周知,亚瑟王是一 ...

  5. bzoj 1419 Red is good - 动态规划 - 概率与期望

    Description 桌面上有R张红牌和B张黑牌,随机打乱顺序后放在桌面上,开始一张一张地翻牌,翻到红牌得到1美元,黑牌则付出1美元.可以随时停止翻牌,在最优策略下平均能得到多少钱. Input 一 ...

  6. Codeforces Round #202 (Div. 1) A. Mafia 推公式 + 二分答案

    http://codeforces.com/problemset/problem/348/A A. Mafia time limit per test 2 seconds memory limit p ...

  7. Codeforces Round #402 (Div. 2) D. String Game(二分答案水题)

    D. String Game time limit per test 2 seconds memory limit per test 512 megabytes input standard inpu ...

  8. Codeforces Round #402 (Div. 2) D题 【字符串二分答案+暴力】

    D. String Game Little Nastya has a hobby, she likes to remove some letters from word, to obtain anot ...

  9. Educational Codeforces Round 80 (Rated for Div. 2)D(二分答案,状压检验)

    这题1<<M为255,可以logN二分答案后,N*M扫一遍表把N行数据转化为一个小于等于255的数字,再255^2检验答案(比扫一遍表复杂度低),复杂度约为N*M*logN #define ...

随机推荐

  1. js中var a=new Object()和var a={}有什么区别吗?

    应该是没有区别的,两者都是生成一个默认的Object对象.js和其它语言一样,一切对象的基类都是Object,所以,new Object()和简易的{}是同样的空对象,就是默认的对象.本来我以为{}应 ...

  2. LA 2218 Triathlon(半平面交)

    Triathlon [题目链接]Triathlon [题目类型]半平面交 &题解: 做了2道了,感觉好像套路,都是二分答案,判断半平面交是否为空. 还有刘汝佳的代码总是写const +& ...

  3. 使用spring的特殊bean完成分散配置

    1.在使用分散配置时,spring的配置文件applicationContext.xml中写法如下: <!-- 引入db.properties文件, --> <context:pro ...

  4. C# Dapper 简单实例

    /// <summary> /// 分页信息 /// </summary> public class PageInfo<T>     {         /// & ...

  5. Maven的特点、优点-功能摘要

    Maven功能摘要 以下是Maven的主要特点: 遵循最佳实践的简单项目设置 - 在几秒钟内启动新项目或模块 所有项目的一致使用 - 意味着新开发人员进入项目的时间不会增加 卓越的依赖管理,包括自动更 ...

  6. react修改app.js添加中文内容后中文部分乱码解决

    [问题]:配置完react后修改app.js内容时添加中文出现如下乱码的中文. [A解决]文档——文本编码——转换文本编码,在弹出窗口修改,确定,搞定 [B解决]首先在EditPlus内:工具——首选 ...

  7. 多线程(threading)示例

    一.多线程简单示例 import threading,time print('第一线程(默认):程序开始啦!') def takeANap(): time.sleep(5) print('第二线程:5 ...

  8. BIOS 搭配 MBR/GPT 的开机流程

    鸟哥私房菜书上内容: BIOS 搭配 MBR/GPT 的开机流程 在计算机概论里面我们有谈到那个可爱的BIOS与CMOS两个东西, CMOS是记录各项硬件参数且嵌入在主板上面的储存器,BIOS则是一个 ...

  9. antd-design model 数据特点

  10. Maven依赖中的scope详解,在eclipse里面用maven install可以编程成功,到服务器上用命令执行报VM crash错误

    Maven依赖中的scope详解 项目中用了<scope>test</scope>在eclipse里面用maven install可以编译成功,到服务器上用命令执行报VM cr ...