Little Pony and Expected Maximum Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 454C Description Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter…

题目链接 题意 : 一个m面的骰子,掷n次,问得到最大值的期望. 思路 : 数学期望,离散时的公式是E(X) = X1*p(X1) + X2*p(X2) + …… + Xn*p(Xn) p(xi)的是所有最大值是xi的情况数/总情况数一共是m^n种,掷n次,所有最大值是xi的情况数应该是xi^n,但是这里边却包含着最大值非xi且不超过xi的种数,所以再减去最大值是xi-1或者最大值不超过这个的情况数.即sum += xi * (xi^n-(xi-1)^n)/m^n,但是这样求肯定是不行,因为m…

A. Little Pony and Expected Maximum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept…

A. Little Pony and Expected Maximum Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/453/problem/A Description Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing.…

题目描述: Little Pony and Expected Maximum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she k…

A. Little Pony and Expected Maximum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept…

LINK:Little Pony and Expected Maximum 容易设出状态f[i][j]表示前i次最大值为j的概率. 转移很显然 不过复杂度很高. 考虑优化.考虑直接求出最大值为j的概率 对于1显然\((\frac{1}{m})^n\) 对于2 显然有\((\frac{2}{m})^n\)不过这其中有一部分是1的概率容斥掉即可. 对于更大的点数也是如此记一个前缀和sum就好了. (不过这个前缀和是不必要的 刚好等于上一次求出的概率 需要小数快速幂 就没了. const ll MAX…

Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game. The dice has m faces: the firs…

http://codeforces.com/problemset/problem/453/A 题目大意: 给定一个m面的筛子,求掷n次后,得到的最大的点数的期望 题解 设f[i]表示掷出 <= i 的点数的概率 ans = sigma{i*(f[i]-f[i-1])} 单个f[i]直接快速幂计算 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typede…

/* 我们枚举每次选择最大数值的情况:m个数, 投掷n次 最大值是1: 1种 2: 2^n-1 3: 3^n-2^n ..... m: m^n-(m-1)^n 所以最后的结果=sum((k/m)^n - ((k-1)/m)^n) (1<=k<=m) 不要这样求(k^n/m^n)数据可能会很大! */ #include<iostream> #include<cstdio> #include<cmath> using namespace std; int mai…

题目链接 题意:一个m个面的骰子,抛掷n次,求这n次里最大值的期望是多少.(看样例就知道) 分析: m个面抛n次的总的情况是m^n, 开始m==1时,只有一种 现在增加m = 2,  则这些情况是新增的那个的第一次的结果的后面最大的都是新增的, 之前的这些的分支也加上这个数,而且这个数是这一支里最大的,也就是说新增产生的结果 全都是m = 2的这个结果,所以用现在的总的情况减去之前的总的情况. 所以: 最大值是1: 1种 2: 2^n-1 3: 3^n-2^n ..... m: m^n-(m-1…

题意:一个筛子有m个面,然后扔n次,求最大值的期望; 思路:最大值为1 有1种,2有2n-1种,  3有3n -2n 种   所以为m的时有mn -(m-1)n 种,所以分别求每一种的概率,然后乘以这个值求和就可以. #include <cstdio> #include <cstring> #include <iostream> #include <cmath> #include <algorithm> using namespace std;…

水了一上午.. 拿6面举例子吧,因为是投掷m次取最大,最大是1概率(1/6)^m;最大是2就可以取到(1,2)那么概率就是(1/3)^m-(1/6)^m.(当前减去上一个) #include<bits/stdc++.h> #define LL long long #define LD long double #define N 100005 using namespace std; inline int ra() { ,f=; char ch=getchar(); ; ch=getchar()…

A. Little Pony and Crystal Mine 水题,每行D的个数为1,3.......n-2,n,n-2,.....3,1,然后打印即可 #include <iostream> #include <vector> #include <algorithm> #include <string> using namespace std; int main(){ int n; cin >> n; vector<string>…

A.Little Pony and Expected Maximum Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 454C Description Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutte…

A .Little Pony and Expected Maximum pro:给定M,N,表示一个M面的骰子,甩N次,问出现的最大的数的期望. sol:容斥,f(i)表示最大数<=i的期望,那么最大数=x的期望就是f(x)-f(x-1); #include<bits/stdc++.h> #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; ; double dp[maxn]; double qpow(doubl…

C. Little Pony and Expected Maximum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept…

一.前言 Wish3D出品的Smart3D系列教程中,前面一讲说明了小物件的照片三维重建,相信大家对建模的流程有了一定的了解.这次讲解中,我们将演示说明以一组无人机倾斜摄影照片为原始数据,通过Smart3D建模软件,重建生成三维地形的过程.与上次不同的是,这次的建模需要对导入的照片预先做一些处理,详情后面会一一说明. 二.工具材料 包括Smart3D建模软件,一组垂直拍摄而且多角度.重叠度满足重建要求的航片.航片对应的pos数据文件 三.方法步骤 关于通过无人机航拍的照片,照片进行三维重建生产模…

点击打开链接 根据TI官方MSP430平台移植修改过来的,在Omap37xx(wince)平台测试,理论上和平台无关,伸手党,赶紧复制粘贴代码吧.如果这篇文章帮助了你,给了好评也无妨. [cpp] view plaincopyprint? /* ================================================================================ *             Texas Instruments OMAP(TM) Platf…

http://electronicdesign.com/digital-ics/digital-current-mode-control-challenges-analog-counterparts Digital Current-Mode Control Challenges Analog Counterparts Electronic Design Bryan Kris Thu, 2006-11-16 (All day) Switch-mode power supplies can use…

Description Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game. The dice has m fac…

Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game. The dice has m faces: the firs…

A real-time operating system (RTOS) for use with minimal-memory controllers has a kernel for managing task execution, including context switching, a plurality of defined tasks, individual ones of the tasks having subroutines callable in nested levels…

网上DS12C887的文章涉及到时间的存储格式使用的都是二进制代码,究竟使用BCD码该如何操作?正好美信官网上有一篇文章.美信官网不稳定,先贴到这里,有时间再翻译. 原文链接 State Machine Logic in Binary-Coded Decimal (BCD)-Formatted Real-Time Clocks 摘要 : When developing code to operate a real-time clock (RTC), it is often beneficial…

Mobileye独创性创新 尽管存在相似之处,但Nvidia的SFF无法与Mobileye的RSS相匹配,后者是领先的AV安全模型 迈向无人驾驶的未来,Mobileye继续以新的创新引领行业,不仅将使全自动驾驶汽车(AVs)成为可能,而且还将使人类驾驶的汽车比以往任何时候都更安全.多年来,取得了许多行业首创:2007年实现了摄像头和雷达融合,2010年实现了行人检测警告,2011年实现了仅摄像头向前碰撞警告,2013年实现了纯摄像头自动巡航控制,2015年的免提协助,2016年的众包高清映射,2…

0,熵的描述 熵(entropy)指的是体系的混沌的程度(可也理解为一个随机变量的不确定性),它在控制论.概率论.数论.天体物理.生命科学等领域都有重要应用,在不同的学科中也有引申出的更为具体的定义,是各领域十分重要的参量.熵由鲁道夫·克劳修斯(Rudolf Clausius)提出,并应用在热力学中.后来在,克劳德·艾尔伍德·香农(Claude Elwood Shannon)第一次将熵的概念引入到信息论中来.----baidu 下面我们将从随机变量开始一步一步慢慢理解熵. 1,随机变量(rand…

https://www.topcoder.com/community/data-science/data-science-tutorials/maximum-flow-augmenting-path-algorithms-comparison/ 存档用... By  Zealint– TopCoder Member Discuss this article in the forums With this article, we’ll revisit the so-called “max-flow…

参考:Fitting a Model by Maximum Likelihood 最大似然估计是用于估计模型参数的,首先我们必须选定一个模型,然后比对有给定的数据集,然后构建一个联合概率函数,因为给定了数据集,所以该函数就是以模型参数为自变量的函数,通过求导我们就能得到使得该函数值(似然值)最大的模型参数了. Maximum-Likelihood Estimation (MLE) is a statistical technique for estimating model parameters…

D. Expected diameter of a tree time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Pasha is a good student and one of MoJaK's best friends. He always have a problem to think about. Today they…

Problem For a random variable XX taking integer values between 1 and nn, the expected value of XX is E(X)=∑nk=1k×Pr(X=k)E(X)=∑k=1nk×Pr(X=k). The expected value offers us a way of taking the long-term average of a random variable over a large number o…