题目链接 题意 :把硬币往棋盘上扔,分别求出硬币占1,2,3,4个格子的时候的概率. 思路 : 求出公式输出,不过要注意输出格式,我还因为输入的时候用了int类型错了好几次..... #include <stdio.h> #include <string.h> #include <iostream> #include <math.h> #define PI acos(-1.0) using namespace std ; int main() { int k…

题目链接 概率问题,像是概率论上学的均匀分布,是不是呢,忘了... 概率同面积有关系,我写的各种搓,然后此题格式十分变态,=前有的时候俩空格,有的时候一个空格.代码各种搓. #include <cstdio> #include <cstring> #include <iostream> #include <cmath> #include <algorithm> using namespace std; #define PI 3.141592653…

这题主要是推导数学公式!!! 将概率问题转化为圆心所在的面积! 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #define pi acos(-1.0) using namespace std; int main() { ,k; double a1,a2…

题目链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1269 10328 - Coin Toss Time limit: 3.000 seconds 问题描述 Toss is an important part of any event. When everything becomes equal toss is the ultim…

/* poj 1474 Video Surveillance - 求多边形有没有核 */ #include <stdio.h> #include<math.h> const double eps=1e-8; const int N=103; struct point { double x,y; }dian[N]; inline bool mo_ee(double x,double y) { double ret=x-y; if(ret<0) ret=-ret; if(ret&…

/* poj 1279 Art Gallery - 求多边形核的面积 */ #include<stdio.h> #include<math.h> #include <algorithm> using namespace std; const double eps=1e-8; struct point { double x,y; }dian[20000+10]; point jiao[203]; struct line { point s,e; double angle;…

ssworld VS DDD Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1487    Accepted Submission(s): 304 Problem Description One day, sssworld and DDD play games together, but there are some special…

Coin Toss Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVA. Original ID: 1032864-bit integer IO format: %lld      Java class name: Main   Toss is an important part of any event. When everything becomes equal toss is the ul…

关于有向图走"无限次"后求概率/期望的口胡/[题解]HNCPC2019H 有向图 全是口胡 假了不管 讨论的都是图\(G=(V,E),|V|=n,|E|=m\)上的情况 "走无限次"这个概念很抽象,严谨的证明以及描述和概率的收敛性有关,由于我也不会在此就不讨论这些,但是根据一些概率的知识,可以发现,其实走无限次可以这样描述: 由于使用概率不好描述在无限次的情况时,每个点和点之间的关系,所以用期望.到时候根据期望的定义式反过来求概率.可能的问题是,"不是走无…

POJ 3080 Blue Jeans (求最长公共字符串) Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. As an IB…