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 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
Sample Output
题意
ACM比赛中,共 M 道题,T 个队, p[i][j] 表示第 i 队解出第 j
思路
首先,每队至少解出一题中一定包含所有队解题数目至少为1并且都是少于 n 的,如果减去这一部分便是至少有一个队伍的解题数目大于 n 了,至于冠军队嘛!我们不用关心,因为我们只需要让它大于 n
- ans: 所有队至少解出一题的概率
- as : 所有队至少解出一题但是不超过 n−1 题的概率最终答案: ans−as
一个队伍解出不超过 n−1 道题目的概率是解出 1、2、...、n−1
pd[j][k] 表示在前 j 道题中做对 k
pd[j][k]=pd[j−1][k]∗(1.0−p[i][j])+pd[j−1][k−1]∗p[i][j]
其中考虑 j<script type="math/tex" id="MathJax-Element-51">j</script> 题,要么做对,要么做错,只有这两种情况,其概率相加。
AC 代码