Check the difficulty of Problems-poj 2151 (probability +dp)

There are T (1<t<=1000) teams and M (0<m<=30) topics, known to each team I solve each topic J's probability p[i][j], now asked: Each team to solve at least one problem, and the most problem-solving team of the number of questions not less than N (0<n& What is the probability of lt;=m?

P[I][J] Indicates the probability of the first team doing the question J. Dp[i][j][k] indicates that the first team has done the probability of K-channel in the first J question.
Dp[i][j][k] = dp[i][j-1][k-1]* (P[i][j]) +dp[i][j-1][k]* (1-p[i][j]);
Find out the probability that each team will do at least 1 questions: Ans1 *= 1-dp[i][m][0];
Find out that each team only do the right 1 ~ n-1 The probability of the problem ans2 namely: (Each team to do 1 ~ n-1 the probability of adding, and the results of each team multiplied);
Then subtract the two ans1-ans2

1#include <stdio.h>2#include <stdlib.h>3#include <string.h>4 intM,t,n;5 Doublep[1002][ +];6 Doubledp[1002][ +][ +];7 Doublesum[1002][ +];8 intMain () {9scanf" %d%d%d",&m,&t,&N);Ten      while(m&&n&&T) { OneMemset (P,0,sizeof(P)); A          for(intI=1; i<=t;i++){ -              for(intj=1; j<=m;j++){ -scanf"%LF",&p[i][j]); the             } -         } -Memset (DP,0,sizeof(DP)); -          for(intI=1; i<=t;i++){ +dp[i][1][1]=p[i][1]; -dp[i][1][0]=1-p[i][1]; +  A         } at          for(intI=1; i<=t;i++){ -              for(intj=2; j<=m;j++){ -                  for(intk=0; k<=j;k++){ -dp[i][j][k]=dp[i][j-1][k]* (1.0-p[i][j]); -                     if(k>0) dp[i][j][k]+=dp[i][j-1][k-1]*P[i][j]; -                 } in             } -         } tomemset (SUM,0,sizeof(sum)); +          for(intI=1; i<=t;i++){ -sum[i][0]=dp[i][m][0]; the              for(intj=1; j<=m;j++) { *sum[i][j]=sum[i][j-1]+Dp[i][m][j]; $             }Panax Notoginseng         } -         Doubleans1=1.0, ans2=1.0; the          for(intI=1; i<=t; i++)  { +ans1*=sum[i][m]-sum[i][0]; AAns2*= (sum[i][n-1]-sum[i][0]); the             } +printf"%.3lf\n", ans1-ans2); -scanf" %d%d%d",&m,&t,&N); $     } $  -     return 0; -}

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5766		Accepted: 2515

Description

Organizing a programming contest is a easy job. To avoid making the problems too difficult, the organizer usually expect the contest result satisfy the following S
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 have studied out the contest problems, and through the result of preliminary contest, the organizer can E Stimate 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-calculate the probability, all of the teams solve at least one problem, and at the same time the Champio N team solves at least n problems?

Input

The input consists of several test cases. The first line of all test case contains three integers m (0 < M <=), T (1 < T <=) and N (0 < N &L T;= 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 is processed.

Output

For each test case, please output the answer-a separate line. The result should is rounded to three digits after the decimal point.

Sample Input

2 2 20.9 0.91 0.90 0 0

Sample Output

0.972

Check the difficulty of Problems-poj 2151 (probability +dp)