"Uva11181" probability | given (Probability

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Question Translation

N people are going to buy things. The probability of buying things is P [I].

Now I know that R people have bought things. In this case, I want to calculate the probability that everyone will buy things.

Thanks @ s_r_f for providing translation

Description

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Input/Output Format

Input Format:

 

 

Output Format:

 

 

Input and Output sample input sample #1: Copy

3 20.100.200.305 10.100.100.100.100.100 0

Output example #1: Copy

Case 1:0.4130430.7391300.847826Case 2:0.2000000.2000000.2000000.2000000.200000

Question

What is this question about ?!!! (Crying

Set $ e $ to event "$ r$ items bought by individuals", $ e_ I $ to event "No. $ I $ items bought by individuals ".

According to the formula, $ P (e_ I | E) = P ({e_ I} e)/P (e) $.

The problem is converted to $ P (e) $ and $ P ({e_ I} e) $.

For $ P (e) $, use $ DFS $ to list whether or not each user has bought it.

Set $ DFS $ to the first $ I $ individual, the person before itCombination of purchased or notThe probability is $ now $,

If you buy $ I $, the probability is $ now * P [I] $; otherwise, $ now * (1-p [I]) $.

Let's just move it over again.

When $ DFS $ arrives at $ N $, and $ r$ is bought,

Then $ P (e) + = now $.

Then, find $ P ({e_ I} e) $.

In fact, it is almost the same, but there is an additional limit of "the $ I $ individual must buy", so force the $ I $ individual to buy at $ DFS $.

1 qwerta 2 uva11181 probability | given accepted 3 code C ++, 0.79kb 4 submission time 18:50:22 5 elapsed time/memory 200 ms, 0kb 6 7 # include <iostream> 8 # include <cstdio> 9 Using namespace STD; 10 double p [23]; 11 double PE; 12 int n, R; 13 void DFS (int x, int D, double now) 14 {15 if (D> r) return; 16 if (x = n + 1) 17 {18 if (D = R) 19 PE + = now; 20 return; 21} 22 // mark23 DFS (x + 1, D + 1, now * P [x]); 24 // dismark25 DFS (x + 1, D, now * (1-p [x]); 26 return; 27} 28 double PEI; 29 void DFSs (int x, int D, int I, double now) 30 {31 if (D> r) return; 32 If (x = n + 1) 33 {34 if (D = r) 35 Pei + = now; 36 return; 37} 38 If (x = I) 39 {DFSs (x + 1, d + 1, I, now * P [x]); return;} 40 // mark41 DFSs (x + 1, D + 1, I, now * P [x]); 42 // dismark43 DFSs (x + 1, D, I, now * (1-p [x]); 44 return; 45} 46 int main () 47 {48 IOs: sync_with_stdio (false); 49 int Tim = 0; 50 while (CIN> N> r) 51 {52 If (n = 0) break; 53 printf ("case % d: \ n", ++ Tim); 54 for (INT I = 1; I <= N; ++ I) 55 CIN> P [I]; 56 Pe = 0; 57 DFS (1, 0); 58 for (INT I = 1; I <= N; ++ I) 59 {60 Pei = 0; 61 DFSs (1, 0, I, 1); 62 printf ("%. 6f \ n ", PEI/PE); 63} 64} 65 return 0; 66}

 

"Uva11181" probability | given (Probability