zoukankan      html  css  js  c++  java
  • zoj3696(泊松分布)

    p(k)=(y^k) / (k!) * e^(-y)

    其中的y就是平均值

    k就是我们要求的大小。

    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

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    #include <algorithm>
    #include <math.h>
    #include <map>
    #include <queue>
    #include <sstream>
    #include <iostream>
    using namespace std;
    #define INF 0x3fffffff
    
    int main()
    {
        //freopen("//home//chen//Desktop//ACM//in.text","r",stdin);
        //freopen("//home//chen//Desktop//ACM//out.text","w",stdout);
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int n;
            double l;
            scanf("%d%lf",&n,&l);
            double sum=0;
            for(int k=0;k<=n;k++)
            {
                double tmp;
                tmp=exp(-l);
                for(int i=k;i>=1;i--)
                {
                    tmp=tmp*l;
                    tmp=tmp/i;
                }
                sum+=tmp;
            }
            printf("%.3lf
    ",sum);
        }
        return 0;
    }

  • 相关阅读:
    php 解析xml
    php
    php 设置自动加载某个页面
    Mac
    mysql
    Git
    C#
    C# 正则表达式
    C# ASCII码排序
    (转)datagridview 自定义列三步走
  • 原文地址:https://www.cnblogs.com/chenhuan001/p/3153075.html
Copyright © 2011-2022 走看看