概率dp ZOJ 3640
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld
& %llu
System Crawler (2014-10-22)
Description
Background
If thou doest well, shalt thou not be accepted? and if thou doest not well, sin lieth at the door. And unto thee shall be his desire, and thou shalt rule over him.
And Cain talked with Abel his brother: and it came to pass, when they were in the field, that Cain rose up against Abel his brother, and slew him.
And the LORD said unto Cain, Where is Abel thy brother? And he said, I know not: Am I my brother's keeper?
And he said, What hast thou done? the voice of thy brother's blood crieth unto me from the ground.
And now art thou cursed from the earth, which hath opened her mouth to receive thy brother's blood from thy hand;
When thou tillest the ground, it shall not henceforth yield unto thee her strength; a fugitive and a vagabond shalt thou be in the earth.
—— Bible Chapter 4
Now Cain is unexpectedly trapped in a cave with N paths. Due to LORD's punishment, all the paths are zigzag and dangerous. The difficulty of the ith path is ci.
Then we define f as the fighting capacity of Cain. Every day, Cain will be sent to one of the N paths randomly.
Suppose Cain is in front of the ith path. He can successfully take ti days to escape from the cave as long as his fighting capacity f is larger than ci. Otherwise, he has to keep trying day after day. However,
if Cain failed to escape, his fighting capacity would increase cias the result of actual combat. (A kindly reminder: Cain will never died.)
As for ti, we can easily draw a conclusion that ti is closely related to ci. Let's use the following function to describe their relationship:

After D days, Cain finally escapes from the cave. Please output the expectation of D.
Input
The input consists of several cases. In each case, two positive integers N and f (n ≤ 100, f ≤ 10000) are given in the first line. The second line includes N positive integers ci (ci ≤ 10000,
1 ≤ i ≤ N)
Output
For each case, you should output the expectation(3 digits after the decimal point).
Sample Input
3 1
1 2 3
Sample Output
6.889
/*************************************************************************
> File Name: t.cpp
> Author: acvcla
> Mail: acvcla@gmail.com
> Created Time: 2014年10月21日 星期二 21时33分55秒
************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn = 20000 + 10;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
int n,c[105],f;
double dp[maxn];
double d(int f){
if(dp[f]>0)return dp[f];
dp[f]=0;
for(int i=1;i<=n;i++){
if(f>c[i]){
int t=(1+sqrt(5))*c[i]*c[i]/2;
dp[f]+=(double)t/n;
}else{
dp[f]+=(1+d(f+c[i]))/n;
}
}
return dp[f];
}
int main(int argc, char const *argv[])
{
while(~scanf("%d%d",&n,&f)){
memset(dp,0,sizeof dp);
for(int i=1;i<=n;i++)scanf("%d",c+i);
printf("%.3f\n",d(f));
}
return 0;
}
概率dp ZOJ 3640的更多相关文章
- [概率dp] ZOJ 3822 Domination
题意: 给N×M的棋盘.每天随机找一个没放过棋子的格子放一个棋子 问使得每一个每列都有棋子的天数期望 思路: dp[i][j][k] 代表放了i个棋子占了j行k列 到达目标状态的期望 然后从 dp[n ...
- zoj 3640 Help Me Escape 概率DP
记忆化搜索+概率DP 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include ...
- zoj 3822(概率dp)
ZOJ Problem Set - 3822 Domination Time Limit: 8 Seconds Memory Limit: 131072 KB Special Ju ...
- zoj 3822 Domination (概率dp 天数期望)
题目链接 参考博客:http://blog.csdn.net/napoleon_acm/article/details/40020297 题意:给定n*m的空棋盘 每一次在上面选择一个空的位置放置一枚 ...
- ZOJ 3822 Domination(概率dp 牡丹江现场赛)
题目链接:problemId=5376">http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5376 Edward ...
- ZOJ 3822 Domination 概率dp 难度:0
Domination Time Limit: 8 Seconds Memory Limit: 131072 KB Special Judge Edward is the headm ...
- zoj 3822 Domination 概率dp 2014牡丹江站D题
Domination Time Limit: 8 Seconds Memory Limit: 131072 KB Special Judge Edward is the headm ...
- ZOJ 3822 ( 2014牡丹江区域赛D题) (概率dp)
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5376 题意:每天往n*m的棋盘上放一颗棋子,求多少天能将棋盘的每行每列都至少有 ...
- 概率dp专场
专题链接 第一题--poj3744 Scout YYF I 链接 (简单题) 算是递推题 如果直接推的话 会TLE 会发现 在两个长距离陷阱中间 很长一部分都是重复的 我用 a表示到达i-2步的概率 ...
随机推荐
- Linux查看系统状态及备份
1. 如何看当前Linux系统有几颗物理CPU和每颗CPU的核数?cat /proc/cpuinfo将CPU的总核数除以物理CPU的个数,得到每颗CPU的核数.2. 查看系统负载有两个常用的命令,是哪 ...
- 后缀自动机/回文自动机/AC自动机/序列自动机----各种自动机(自冻鸡) 题目泛做
题目1 BZOJ 3676 APIO2014 回文串 算法讨论: cnt表示回文自动机上每个结点回文串出现的次数.这是回文自动机的定义考查题. #include <cstdlib> #in ...
- poj1159 dp(滚动数组优化)
H - 简单dp 例题扩展 Crawling in process... Crawling failed Time Limit:3000MS Memory Limit:65536KB ...
- C++智能指针初学小结
本篇随笔仅作为个人学习<C++ Primer>智能指针一节后的部分小结,抄书严重,伴随个人理解.主要介绍shared_ptr.make_shared.weak_ptr的用法和联系. C++ ...
- android中细节效果总结
android中细节效果总结 andorid取消最上方的标题同时全屏显示 Source code protected void onCreate(Bundle savedInstanceS ...
- js页面加载进度条(这个就比较正式了,改改时间就完事儿)
不废话,直接上代码 思路不难,就是一个animate方法配合随机数 duration内个三秒钟,是自定义的,可以改成页面加载时间,这样就完美了 <!doctype html> <ht ...
- 浅析Javascript原型继承(转)
引自: http://blog.csdn.net/kittyjie/article/details/4380918 原作者解释的浅显易懂,非常不错的JavaScript prototype总结 JS没 ...
- cubieboardtruck安装
1.命令关闭所有led灯 ls /sys/class/leds/*/brightness | xargs -i -n1 echo "echo 0 > {}" | sh 如果需 ...
- 使用Japserreport填充报表数据(3)
E中以PDF文件的格式显示静态的中文字符串,在大多数的情况下,打印的数据来自于一些变量,在JasperReports工具中传递数据并填充到 报表只有两种方式,即使用Parameters参数和JRDat ...
- nyoj-366-D的小L(求全排列)
D的小L 时间限制:4000 ms | 内存限制:65535 KB 难度:2 描述 一天TC的匡匡找ACM的小L玩三国杀,但是这会小L忙着哩,不想和匡匡玩但又怕匡匡生气,这时小L给匡匡 ...