????題干：

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 0.9 0.9 1 0.9 0 0 0

Sample Output

0.972

題目大意：

ACM比賽中，共M道題，T個(gè)隊(duì)，pij表示第i隊(duì)解出第j題的概率，問(wèn) 每隊(duì)至少解出一題且冠軍隊(duì)至少解出N道題的概率。

解題報(bào)告：

假設(shè)ans1為每個(gè)隊(duì)至少解出一題的概率。

假設(shè)ans2為每個(gè)隊(duì)解出的題數(shù)均在 [1.n-1] 之間的概率。

這樣最終答案就是ans1-ans2。

ans1可以在讀入的數(shù)據(jù)中預(yù)處理出來(lái)，ans2通過(guò)dp求解。

dp[i][j][k]表示第i個(gè)隊(duì)在第j個(gè)題中解出k個(gè)題的概率。
求出后累加一下就好了。注意別忘dp[][j][0]需要單獨(dú)處理。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int m,t,n;//m<=30,t<=1000 double p[1005][33]; double dp[1005][33][33];//dp[i][j][k]第i個(gè)隊(duì)在第j個(gè)題中解出k個(gè)題的概率 double po,ans1,ans2;//ans1是所有人都做出一個(gè)題的概率 int main() {while(~scanf("%d%d%d",&m,&t,&n) && m+t+n) {ans1=ans2=1;for(int i = 1; i<=t; i++) {po=1;for(int j = 1; j<=m; j++) scanf("%lf",&p[i][j]),po *= (1-p[i][j]);ans1 *= (1-po);}for(int i = 1; i<=t; i++) {dp[i][0][0]=1;for(int j = 1; j<=m; j++) {dp[i][j][0] = dp[i][j-1][0]*(1-p[i][j]);for(int k = 1; k<=m; k++) dp[i][j][k] = dp[i][j-1][k]*(1-p[i][j]) + dp[i][j-1][k-1]*p[i][j];}}for(int i = 1; i<=t; i++) {po=0;for(int j = 1; j<=n-1; j++) po += dp[i][m][j];ans2 *= po;}printf("%.3f\n",ans1-ans2);}return 0 ; }

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