POJ2096 Collecting Bugs

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2019寒假集训题

A.Collecting Bugs

（Html Address:http://poj.org/problem?id=2096）

题面

Ivan is fond of collecting. Unlike other people who collect post stamps,coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers exactly one bug in the program and adds information about it and its category into a spreadsheet. When he finds bugs in all bug categories, he calls the program disgusting, publishes this spreadsheet on his home page, and forgets completely about the program.

Two companies, Macrosoft and Microhard are in tight competition. Microhard wants to decrease sales of one Macrosoft program. They hire Ivan to prove that the program in question is disgusting. However, Ivan has a complicated problem. This new program has s subcomponents, and finding bugs of all types in each subcomponent would take too long before the target could be reached. So Ivan and Microhard agreed to use a simpler criteria —- Ivan should find at least one bug in each subsystem and at least one bug of each category.

Macrosoft knows about these plans and it wants to estimate the time that is required for Ivan to call its program disgusting. It’s important because the company releases a new version soon, so it can correct its plans and release it quicker. Nobody would be interested in Ivan’s opinion about the reliability of the obsolete version.

A bug found in the program can be of any category with equal probability. Similarly, the bug can be found in any given subsystem with equal probability. Any particular bug cannot belong to two different categories or happen simultaneously in two different subsystems. The number of bugs in the program is almost infinite, so the probability of finding a new bug of some category in some subsystem does not reduce after finding any number of bugs of that category in that subsystem.

Find an average time (in days of Ivan’s work) required to name the program disgusting.

简要翻译

一个软件有$s$个子系统，会产生$n$种$bug$。

你每天会发现一个$bug$，这个$bug$属于某个子系统，属于某种$bug$。

每天发现的$bug$，出现在每个子系统中的概率都是$\frac 1s$，属于每种$bug$的概率均为$\frac 1n$。

求找出所有的$n$种$bug$，且每个子系统都找到$bug$所需要的天数的期望。

Input

输入两个整数$n,s$。

Output

输出期望，保留小数点后四位小数。

Sample Input

1 2

Sample Output

3.0000

Hint

$1<n,s\leq 1,000$

设$f(i,j)$表示已经找到$i$种$bug$，存在于$j$个子系统中的期望还要多少天。

那么$f(n,s)=0$，$f(0,0)$即为~~（正义）~~答案。

状态转移如下：

$1.$发现了新$bug$种类，存在于新子系统，$p_1=\frac{n-i}n×\frac{s-j}s$，还需要$f(i+1,j+1)$天。

$2.$发现了新$bug$种类，存在于旧子系统，$p_2=\frac{n-i}n×\frac js$，还需要$f(i+1,j)$天。

$3.$发现了旧$bug$种类，存在于新子系统，$p_3=\frac in×\frac {s-j}s$，还需要$f(i,j+1)$天。

$4.$发现了旧$bug$种类，存在于旧子系统，$p_4=\frac in×\frac js$，还需要$f(i,j)$天。

所以期望的转移方程为：

$f(i,j)=p_1f(i+1,j+1)+p_2f(i+1,j)+p_3f(i,j+1)+p_4f(i,j)+1$

移项$f(i,j)$可得：

$f(i,j)=\frac{p_!f(i+1,j+1)+p_2f(i+1,j)+p_3f(i,j+1)+1}{1-p_4}$

时间复杂度为$O(ns)$。

Code

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int read()
{
	int x = 0, w = 0;
	char ch = 0;
	while (!isdigit(ch)) {w |= ch == '-'; ch = getchar();}
	while (isdigit(ch)) {x = (x << 3) + (x << 1) + ch - '0'; ch = getchar();}
	return w ? -x : x;
}
int n, s;
double f[1010][1010];
int main(int argc, char const *argv[])
{
	while(~scanf("%d%d",&n,&s))
	{
		memset(f,0,sizeof(f));
		for (int i = n; ~i; --i)
			for (int j = s; ~j; --j)
			{
				if (i == n && j == s) continue;
				double p1 = 1.0 * (n - i) / n * (s - j) / s;
				double p2 = 1.0 * (n - i) / n * j / s;
				double p3 = 1.0 * i / n * (s - j) / s;
				double p4 = 1.0 * i / n * j / s;
				f[i][j] = 1.0 * (p1 * f[i + 1][j + 1] + p2 * f[i + 1][j] + p3 * f[i][j + 1] + 1) / (1.0 - p4);
			}
		printf("%.4f\n", f[0][0]);
	}
	return 0;
}

poj有毒啦！

最后明明应该输出“$\%.4lf$”，结果强行”$\%.4f$”才过。

[○･｀Д´･ ○]！！！