Check the difficulty of problems(POJ 2151)
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:
1. All of the teams solve at least one problem. 2. 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 Pij (1 <= i <= T, 1<= j <= 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 <= 30), T (1 < T <= 1000) and N (0 < N <= 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 Pij. 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 Sample Output 0.972 Source POJ Monthly,鲁小石
概率dp
开始理解错了题意,弄了半天也没搞出来,后来理解完了啦,发现概率都忘了。
然后补概率。
设A = “所有队都至少做完一题”, B = “至少存在一个队做完不少于n题”;
P(A) = P(A(B + !B)) = P(AB) + P(A!B);
P(AB) = P(A) - P(A!B);
#include <cstdio> |
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