题意: N个礼品箱, 每个礼品箱内的礼品只有第一个抽到的人能拿到. M个小孩每个人依次随机抽取一个,  求送出礼品数量的期望值. 1 ≤ N, M ≤ 100, 000

挺水的说..设f(x)表示前x个人都选择完成后礼品剩下数的期望值( f(0) = N ), 那么f(x) = f(x - 1) - f(x - 1) / N = f(x - 1) * (N - 1) / N (显然). 那么答案就是等于 N - N * [(N - 1) / N]^M. 后面部分可以用快速幂优化.时间复杂度O(log M). 数据这么小不用快速幂O(M)应该也能过...

-----------------------------------------------------------------------------

#include<bits/stdc++.h>
 
using namespace std;
 
double POW(double p, int a) {
double ans = 1;
for(; a; a >>= 1) {
if(a & 1) ans *= p;
p *= p;
}
return ans;
}
 
int main() {
int N, M;
scanf("%d%d", &N, &M);
printf("%.9lf\n", N - POW(1.0 * (N - 1) / N, M) * N);
return 0;
}

-----------------------------------------------------------------------------

495. Kids and Prizes

Time limit per test: 0.25 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

ICPC (International Cardboard Producing Company) is in the business of producing cardboard boxes. Recently the company organized a contest for kids for the best design of a cardboard box and selected M winners. There are N prizes for the winners, each one carefully packed in a cardboard box (made by the ICPC, of course). The awarding process will be as follows:

  • All the boxes with prizes will be stored in a separate room.
  • The winners will enter the room, one at a time.
  • Each winner selects one of the boxes.
  • The selected box is opened by a representative of the organizing committee.
  • If the box contains a prize, the winner takes it.
  • If the box is empty (because the same box has already been selected by one or more previous winners), the winner will instead get a certificate printed on a sheet of excellent cardboard (made by ICPC, of course).
  • Whether there is a prize or not, the box is re-sealed and returned to the room.

The management of the company would like to know how many prizes will be given by the above process. It is assumed that each winner picks a box at random and that all boxes are equally likely to be picked. Compute the mathematical expectation of the number of prizes given (the certificates are not counted as prizes, of course).

Input

The first and only line of the input file contains the values of N and M ().

Output

The first and only line of the output file should contain a single real number: the expected number of prizes given out. The answer is accepted as correct if either the absolute or the relative error is less than or equal to 10-9.

Example(s)
sample input
sample output
5 7 
3.951424 

sample input
sample output
4 3 
2.3125 

SGU 495. Kids and Prizes( 数学期望 )的更多相关文章

  1. SGU 495 Kids and Prizes:期望dp / 概率dp / 推公式

    题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=495 题意: 有n个礼物盒,m个人. 最开始每个礼物盒中都有一个礼物. m个人依次随 ...

  2. sgu 495. Kids and Prizes (简单概率dp 正推求期望)

    题目链接 495. Kids and Prizes Time limit per test: 0.25 second(s)Memory limit: 262144 kilobytes input: s ...

  3. SGU 495. Kids and Prizes

    水概率....SGU里难得的水题.... 495. Kids and Prizes Time limit per test: 0.5 second(s)Memory limit: 262144 kil ...

  4. 495. Kids and Prizes

    http://acm.sgu.ru/problem.php?contest=0&problem=495 学习:当一条路走不通,换一种对象考虑,还有考虑对立面. 495. Kids and Pr ...

  5. 【SGU】495. Kids and Prizes

    http://acm.sgu.ru/problem.php?contest=0&problem=495 题意:N个箱子M个人,初始N个箱子都有一个礼物,M个人依次等概率取一个箱子,如果有礼物则 ...

  6. SGU495Kids and Prizes 数学期望

    题意: 有n个奖品,m个人排队来选礼物,对于每个人,他打开的盒子,可能有礼物,也有可能已经被之前的人取走了,然后把盒子放回原处.为最后m个人取走礼物的期望. 题解: 本道题与之前的一些期望 DP 题目 ...

  7. Kids and Prizes(SGU 495)

    495. Kids and Prizes Time limit per test: 0.25 second(s)Memory limit: 262144 kilobytes input: standa ...

  8. SGU495Kids and Prizes(数学期望||概率DP||公式)

    495. Kids and Prizes Time limit per test: 0.25 second(s) Memory limit: 262144 kilobytes input: stand ...

  9. [SGU495] Kids and Prizes (概率dp)

    题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=495 题目大意:有N个盒子,里面都放着礼物,M个人依次去选择盒子,每人仅能选一次,如 ...

随机推荐

  1. xen虚拟机操作整理

    1,登陆物理机器 2,查看物理机建立虚拟机的列表 root:~ # xm li Name ID Mem VCPUs State Time(s) Domain-0 0 49450 8 r----- 52 ...

  2. .c和.h文件的区别(头文件与之实现文件的的关系~ )

     .c和.h文件的区别 一个简单的问题:.c和.h文件的区别 学了几个月的C语言,反而觉得越来越不懂了.同样是子程序,可以定义在.c文件中,也可以定义在.h文件中,那这两个文件到底在用法上有什么区别呢 ...

  3. Orchard站点性能优化-预热

    Orchard CMS 包含一个 Warmup Module 模块,当我们的站点在共享主机上的时候,它可以显著的帮助我们快速响应用户访问请求.当你开启这个模块以后,你设置的URL的里面的内容会缓存起来 ...

  4. LightOJ 1317

    Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %lluDescription You probab ...

  5. packstack安装以及centos源配置注意事项

    On CentOS:安装分为四步: 1,$ sudo yum install -y centos-release-openstack-mitaka 2,$ sudo yum update -y 3,$ ...

  6. 字符串匹配算法(KMP)

    字符串匹配运用很广泛,举个简单例子,我们每天登QQ时输入账号和密码,大家有没有想过账号和密码是怎样匹配的呢?登录需要多长时间和匹配算法的效率有直接的关系. 首先理解一下前缀和后缀的概念: 给出一个问题 ...

  7. Java map取value最大值和最小值

    /** * 求Map<K,V>中Value(值)的最小值 * * @param map * @return */ public static Object getMinValue(Map& ...

  8. HTML+CSS笔记 表格,超链接,图片,表单

    表格 给表格加入CSS样式,添加表格边框 语法: <style type="text/css"> table tr td,th{border:1px solid #00 ...

  9. document.domain与js跨域的问题

    以前如果要使iframe里面的脚本能访问parent的内容,但iframe和parent的二级域名相同,那一般都会在两者都写上document.domain="xxx.com" 以 ...

  10. Windows Phone 8初学者开发—第1部分:系列介绍

    原文 Windows Phone 8初学者开发—第1部分:系列介绍 您好,欢迎来到这个包含35课为Window Phone 8平台创建应用程序的系列教程.我叫Bob Tabor,在过去的11年中我一直 ...