Alien's Organ


Time Limit: 2 Seconds      Memory Limit: 65536 KB


There's an alien whose name is Marjar. It is an universal solder came from planet Highrich a long time ago.

Marjar is a strange alien. It needs to generate new organs(body parts) to fight. The generated organs will provide power to Marjar and then it will disappear. To fight for problem of
moral integrity decay on our earth, it will randomly generate new fighting organs all the time, no matter day or night, no matter rain or shine. Averagely, it will generate λ new fighting organs every day.

Marjar's fighting story is well known to people on earth. So can you help to calculate the possibility of that Marjar generates no more than N organs in one day?

Input

The first line contains a single integer T (0
≤ T ≤ 10000), indicating there are T cases
in total. Then the following T lines each contains one integer N (1
≤ N ≤ 100) and one float number λ (1
≤ λ ≤ 100), which are described in problem statement.

Output

For each case, output the possibility described in problem statement, rounded to 3 decimal points.

Sample Input

3
5 8.000
8 5.000
2 4.910

Sample Output

0.191
0.932
0.132

——————————————————————————————————————

输入N和入,入表示平均每天产生多少个物品,求产生物品在N以内的概率是多少?

思路:带入泊松分布定理即可

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <cctype>
#include <sstream>
#include <climits>
#include <unordered_map> using namespace std; #define LL long long
const int INF=0x3f3f3f3f; const double e=exp(1);
int n;
double p; int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%lf",&n,&p);
double ans=pow(e,-p);
double ans1=1;
double k=p;
for(int i=2;i<=n+1;i++)
{
ans1+=k;
k=k*p/i;
}
ans=ans*ans1;
printf("%.3f\n",ans);
}
return 0;
}
Alien's Organ


Time Limit: 2 Seconds      Memory Limit: 65536 KB


There's an alien whose name is Marjar. It is an universal solder came from planet Highrich a long time ago.

Marjar is a strange alien. It needs to generate new organs(body parts) to fight. The generated organs will provide power to Marjar and then it will disappear. To fight for problem of
moral integrity decay on our earth, it will randomly generate new fighting organs all the time, no matter day or night, no matter rain or shine. Averagely, it will generate λ new fighting organs every day.

Marjar's fighting story is well known to people on earth. So can you help to calculate the possibility of that Marjar generates no more than N organs in one day?

Input

The first line contains a single integer T (0
≤ T ≤ 10000), indicating there are T cases
in total. Then the following T lines each contains one integer N (1
≤ N ≤ 100) and one float number λ (1
≤ λ ≤ 100), which are described in problem statement.

Output

For each case, output the possibility described in problem statement, rounded to 3 decimal points.

Sample Input

3
5 8.000
8 5.000
2 4.910

Sample Output

0.191
0.932
0.132

ZOJ3696 Alien's Organ 2017-04-06 23:16 51人阅读 评论(0) 收藏的更多相关文章

  1. HDU4081 Qin Shi Huang's National Road System 2017-05-10 23:16 41人阅读 评论(0) 收藏

    Qin Shi Huang's National Road System                                                                 ...

  2. 团体程序设计天梯赛L1-027 出租 2017-03-23 23:16 40人阅读 评论(0) 收藏

    L1-027. 出租 时间限制 400 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 陈越 下面是新浪微博上曾经很火的一张图: 一时间网上一片求救声, ...

  3. ZOJ2482 IP Address 2017-04-18 23:11 44人阅读 评论(0) 收藏

    IP Address Time Limit: 2 Seconds      Memory Limit: 65536 KB Suppose you are reading byte streams fr ...

  4. 动态链接库(DLL) 分类: c/c++ 2015-01-04 23:30 423人阅读 评论(0) 收藏

    动态链接库:我们经常把常用的代码制作成一个可执行模块供其他可执行文件调用,这样的模块称为链接库,分为动态链接库和静态链接库. 对于静态链接库,LIB包含具体实现代码且会被包含进EXE中,导致文件过大, ...

  5. NYOJ-235 zb的生日 AC 分类: NYOJ 2013-12-30 23:10 183人阅读 评论(0) 收藏

    DFS算法: #include<stdio.h> #include<math.h> void find(int k,int w); int num[23]={0}; int m ...

  6. HDU 2034 人见人爱A-B 分类: ACM 2015-06-23 23:42 9人阅读 评论(0) 收藏

    人见人爱A-B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Su ...

  7. HDU 2035 人见人爱A^B 分类: ACM 2015-06-22 23:54 9人阅读 评论(0) 收藏

    人见人爱A^B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Su ...

  8. HDU2033 人见人爱A+B 分类: ACM 2015-06-21 23:05 13人阅读 评论(0) 收藏

    人见人爱A+B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Su ...

  9. 认识C++中的临时对象temporary object 分类: C/C++ 2015-05-11 23:20 137人阅读 评论(0) 收藏

    C++中临时对象又称无名对象.临时对象主要出现在如下场景. 1.建立一个没有命名的非堆(non-heap)对象,也就是无名对象时,会产生临时对象. Integer inte= Integer(5); ...

随机推荐

  1. 【linux】U盘安装启动出现press the enter key to begin the installation process 就不动弹了

    今天在物理机上安装centOS6.5  64bit 系统的时候,出现了U盘安装启动出现press the enter key to begin the installation process 就不动 ...

  2. jsp页面的基本语法

    JSP全称Java Server Pages,顾名思义就是运行中java服务器中页面,也就是在我们JavaWeb中的动态页面,其本质就是一个Servlet. 学习jsp的基本语法主要就是学习服务器是如 ...

  3. bootstrap全局css样式

    以下从官网抄来的,感觉还是很实用的,运用得好,灵活运用,非常方便快捷,能大大提高开发效率,也为调整不同尺寸的屏幕节省了时间. hidden-xs @media (max-width: 767px){ ...

  4. Java 经典练习题_Day010

    final 变量能被显式地初始化并且只能初始化一次.被声明为 final 的对象的引用不能指向不同的对象.但是 final 对象里的数据可以被改变.也就是说 final 对象的引用不能改变,但是里面的 ...

  5. js复制URL链接

    html: <div style="height:0px; text-indent:-10000px;"><span id="hdcopyurl&quo ...

  6. 第七章 : Git 介绍 (下)[Learn Android Studio 汉化教程]

    Learn Android Studio 汉化教程 Let’s reset even further to remove all traces of your work on the deprecat ...

  7. leetcode69

    public class Solution { public int MySqrt(int x) { long r = x; while (r * r > x) r = (r + x / r) ...

  8. spring错误处理 Build path is incomplete. Cannot find class file for org.springframework.aop.Advisor

    Build path is incomplete. Cannot find class file for org.springframework.aop.Advisor 初学spring,记录一下出现 ...

  9. HashMap,Hash优化与高效散列

    OverView Hash table based implementation of the Map interface. This implementation provides all of t ...

  10. Dubbo Overview

    Overview Architecture Provider: 暴露服务的服务提供方. Consumer: 调用远程服务的服务消费方. Registry: 服务注册与发现的注册中心. Monitor: ...