Poj 2151 check the difficulty of problems (check the difficulty of the problem)

Poj 2151 check the difficulty of problems

Http://poj.org/problem? Id = 2151

Question: at this moment, there are tn questions and DN teams, who know the probability that each team can answer each question. Q: The champion should at least answer n (1 <= n <= Tn) the probability that other teams should answer at least one question

DP + Probability

Method:

F [I] [J] probability of Team I to answer question J

G [I] [J] [k] probability that the first J question of Team I will be used to answer K questions

State transition equation: G [I] [J] [k] = G [I] [J-1] [k-1] * f [I] [J] + G [I] [J-1] [k] * (1-F [I] [J])

 

1/* 2 problem: 2151 User: bibier 3 memory: 5120 K time: 47 Ms 4 Language: G ++ result: accepted 5 */6 7 # include <stdio. h> 8 # include <string. h> 9 # include <iostream> 10 # include <algorithm> 11 # include <cstdio> 12 # include <cstring> 13 # include <cmath> 14 # include <stack> 15 # include <queue> 16 # include <vector> 17 using namespace STD; 18 # define M 0x0f0f0f19 # define min (A, B) (A> B? B: a) 20 # define max (A, B) (A> B? A: B) 21 22 23 float f [1003] [33]; // F [I] [J] the probability that team I performs the J question 24 float G [1003] [33] [33]; // G [I] [J] [k] the probability that the first J question of Team I has a K question is 25 26 // G [I] [J] [k] = G [I] [J-1] [k-1] * f [I] [J] + G [I] [J-1] [k] * (1-F [I] [J]) 27 int main () 28 {29 int TN, dn, N; 30 int I, j, k; 31 While (scanf ("% d", & TN, & DN, & N )! = EOF) 32 {33 If (Tn = 0 & DN = 0 & n = 0) 34 break; 35 memset (G, 0, sizeof (g); 36 memset (F, 0, sizeof (f); 37 38 for (I = 1; I <= DN; I ++) 39 For (j = 1; j <= tn; j ++) 40 scanf ("% F", & F [I] [J]); 41 42 for (I = 1; I <= DN; I ++) 43 {44g [I] [1] [0] = 1-F [I] [1]; 45g [I] [1] [1] = f [I] [1]; 46} 47 48 for (I = 1; I <= DN; I ++) 49 for (j = 2; j <= tn; j ++) 50 for (k = 0; k <= J; k ++) 51 {52 If (k = J) 53g [I] [J] [k] = G [I] [J-1] [k-1] * f [I] [J]; 54 else if (k = 0) 55g [I] [J] [k] = G [I] [J-1] [k] * (1-F [I] [J]); 56 else57 G [I] [J] [k] = G [I] [J-1] [k-1] * f [I] [J] + G [I] [J-1] [k] * (1-F [I] [J]); 58} 59 float all = 1; 60 for (I = 1; I <= DN; I ++) // The final probability for each team to do at least one question is 61 all * = (1-g [I] [tn] [0]); 62 63 float midall = 1; 64 for (I = 1; I <= DN; I ++) 65 {66 float everyd = 0; 67 for (j = 1; j <n; j ++) 68 {69 everyd + = G [I] [tn] [J]; 70} 71 midall * = everyd; 72} 73 all-= midall; 74 printf ("%. 3f \ n ", all); 75} 76 return 0; 77}

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