[POJ2096]Collecting Bugs

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.

Input file contains two integer numbers, n and s (0 < n, s <= 1 000).

Output the expectation of the Ivan‘s working days needed to call the program disgusting, accurate to 4 digits after the decimal point.

1 2

3.0000

题解:
这本来是一道概率期望的入门题
按照高中生物遗传题那样推一下式子,就不难发现有4种情况:
1°在新的子项目里找到新种类病毒
2°在新的子项目里找到旧种类病毒
3°在旧的子项目里找到新种类病毒
4°在旧的子项目里找到旧种类病毒
这样就可以写出递推的式子了
可以O(n*s)写for循环,也可以记忆化递归
但是,
poj输出实数要用%f!!%f!!%f!!

(我会告诉你因为这个卡了我10遍吗)(明明是我先的)

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<iomanip>
 5 using namespace std;
 6 const int N=1100;
 7 int n,s;
 8 double f[N][N];
 9 double k1,k2,k3,k4;
10 int main()
11 {
12     while(scanf("%d%d",&n,&s)==2)
13     {
14         memset(f,0,sizeof(f));
15         for(int i=n;i>=0;i--)
16             for(int j=s;j>=0;j--)
17             {
18                 if(i==n&&j==s)continue;
19                 k1=(n-i)*j*1.0;
20                 k2=(n-i)*(s-j)*1.0;
21                 k3=i*(s-j)*1.0;
22                 k4=(n*s-i*j)*1.0;
23                 f[i][j]=(k1*f[i+1][j]+k2*f[i+1][j+1]+k3*f[i][j+1]+n*s*1.0)/k4;
24             }
25         printf("%.4f\n",f[0][0]);//poj上这里必须写%f 
26     }
27 }

#include<cstdio>
#include<cstring>
using namespace std;
const int N=1010;
int n,s;
double f[N][N];
double dp(int i,int j)
{
    if(i==n&&j==s)return f[i][j]=0;
    if(f[i][j])return f[i][j];
    double tmp=1;
    if(j&&i<n)tmp+=(n-i)*1.0*j/(n*s)*dp(i+1,j);
    if(i<n&&j<s)tmp+=(n-i)*1.0*(s-j)/(n*s)*dp(i+1,j+1);
    if(i&&j<s)tmp+=i*(s-j)/(1.0*n*s)*dp(i,j+1);
    f[i][j]=tmp/(1.0-i*j*1.0/(n*s));
    return f[i][j];
}
int main()
{
    while(scanf("%d%d",&n,&s)==2)
    {
        memset(f,0,sizeof(f));
        printf("%.4f",dp(0,0));
    }
}