POJ 2096 Collecting Bugs용 DP

Description

    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

Input file contains two integer numbers, n and s .

Output

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

Sample Input

1 2

Sample Output

3.0000

Hint

0

About

  s개의 시스템, n가지 유형의 버그가 있는데 매일 하나의 버그를 발견할 수 있다. 하나의 버그는 하나의 유형과 하나의 시스템에 속하고 모든 버그가 특정한 서브시스템에 속하는 확률은 1/s이며 특정한 분류에 속하는 확률은 1/n이다. s개의 시스템과 n가지 버그에 대한 기대를 찾아라.

Solution

우리는 DP[i][j]를 정의하여 i종의 버그를 발견하고 j개의 시스템에 대한 기대를 나타낼 수 있다. 여기서 강조하자면 기대는 하나의 값이다. 네 문제가 확정된 후에 너의 기대도 확정된다. 그것이'변수'라고 생각하지 마라. 내가 인용부호를 쳤으니 너는 나의 뜻을 알아야 한다.DP 방정식을 생각해 봅시다.
DP[i][j]=DP[i][j]∗(i / n)∗(j / s) +
DP[i+1][j]∗(1−i / n)∗(j / s) +
DP[i][j+1]∗(i / n)∗(1−j / n)+
DP[i+1][j+1]∗(1−i / n)∗(1−j / s)+1.
정리해 봅시다.
p1=(i / n)∗(j / s)
p2=(1−i / n)∗(j / s)
p3=(i / n)∗(1−j / n)
p4=(1−i / n)∗(1−j / s)
DP[i][j]=DP[i+1][j]∗p2 +DP[i][j+1]∗p3+DP[i+1][j+1]∗p4+11−p1

Code :

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define clr(a) memset(a, 0, sizeof(a))
#define D double
using namespace std;

inline int read() {
    int i = 0, f = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') f = -1; ch = getchar();
    }
    while(isdigit(ch)) {
        i = (i << 3) + (i << 1) + ch - '0'; ch = getchar();
    }
    return i * f;
}

const int MAXN = 1000 + 5;
D dp[MAXN][MAXN];

int main() {
    int n = read(), s = read();
    for(int i = n; i >= 0; --i)
        for(int j = s; j >= 0; --j) {
            if(i == n && j == s) continue;
            D p1 = 1.0 * i / n * j / s,
            p2 = 1.0 * (n - i) / n * j / s,
            p3 = 1.0 * (n -  i) / n * (s - j) / s,
            p4 = 1.0 * i / n * (s - j) / s;
            dp[i][j] = (dp[i + 1][j] * p2 + dp[i + 1][j + 1] * p3 + dp[i][j + 1] * p4 + 1.0) / (D)(1.0 - p1);
        }
    printf("%.4lf", dp[0][0]);
}