POJ2151 Check the difficulty of problems
Description
Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually expect the contest result satisfy the following two terms:
- All of the teams solve at least one problem.
- The champion (One of those teams that solve the most problems) solves at least a certain number of problems.
Now the organizer has studied out the contest problems, and through the result of preliminary contest, the organizer can estimate the probability that a certain team can successfully solve a certain problem.
Given the number of contest problems M, the number of teams T, and the number of problems N that the organizer expect the champion solve at least. We also assume that team i solves problem j with the probability \(P_{ij}(1 \leqslant i \leqslant T, 1\leqslant j \leqslant M)\). Well, can you calculate the probability that all of the teams solve at least one problem, and at the same time the champion team solves at least N problems?
Input
The input consists of several test cases. The first line of each test case contains three integers \(M (0 < M \leqslant 30), T (1 < T \leqslant 1000)\) and \(N (0 < N \leqslant M)\). Each of the following T lines contains M floating-point numbers in the range of [0,1]. In these T lines, the j-th number in the i-th line is just \(P_{ij}\). A test case of M = T = N = 0 indicates the end of input, and should not be processed.
Output
For each test case, please output the answer in a separate line. The result should be rounded to three digits after the decimal point.
Sample Input
2 2 2
0.9 0.9
1 0.9
0 0 0
Sample Output
0.972
题目大意,给T个队伍,M个题目,给出每个队伍做出每个题的概率\(P_{ij}\),求每个队伍至少做出一道题,且冠军队伍至少做出\(N\)道题的概率
由于每个队伍做题的概率是独立的,所以我们可以提前统计出第\(i\)个队伍至少做出\(k\)个题的概率\(F[i][k]\)
然后我们统计出每个队伍至少做出1道题的概率\(Ans1\),再统计出每个队伍做出\(1~N-1\)的概率\(Ans2\),答案即为\(Ans1-Ans2\)
/*program from Wolfycz*/
#include