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

Sample Output

Source

POJ Monthly,鲁小石
自己yy了很久的算法 写着写着才发现有错 qwq
dp[i][j][k]表示第i个人前j道题目做出k题的概率
s[i][j]表示第i个人 做题数小于等于j的概率
那么设p1=所有人的(1-s[i][0])的乘积
那么p2=所有人的s[i][n-1]-s[i][0]的乘积 发现p1-p2就是答案.. 正解就是那么简单..

 


elijahqi

退役了 现在在商院 偶尔打CF,有时有ACM regional也去玩一下

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