Check the difficulty of problems 概率dp，概率知识很重要
Problem 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?
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
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概率dp
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1 /* 2 dp[i][j][k]代表第i个队在前j道题时解决k道题时的概率 3 p[i][j]代表第i个队解决第j道题时的概率 4 s[i][j]代表第i个队解决0-j道题的概率 5 注意要初始化dp[i][0][0]=1; 6 因为这是必然事件 7 要求有t个队，m道题，每队至少一题，且冠军队不少于n个题的事件概率 8 此时对冠军队不少于n个题，求对立事p2更容易， 9 每队至少一题直接求p1，两者相减就是要求的概率。 10 这道题说明了概率很重要。 11 */ 12 #include<iostream> 13 #include<string> 14 #include<cstring> 15 #include<cstdio> 16 #include<cmath> 17 using namespace std; 18 double dp[1005][50][50],p[1005][50],s[1005][50]; 19 int n,m,t; 20 int i,j,k; 21 double p1,p2; 22 int main() 23 { 24 while(scanf("%d%d%d",&m,&t,&n)!=EOF) 25 { 26 if(!m&&!n&&!t) 27 break; 28 memset(dp,0,sizeof(dp)); 29 memset(p,0,sizeof(p)); 30 memset(s,0,sizeof(s)); 31 for(i=1;i<=t;i++) 32 for(j=1;j<=m;j++) 33 scanf("%lf",&p[i][j]); 34 for(i=1;i<=t;i++) 35 { 36 dp[i][0][0]=1.0; 37 for(j=1;j<=m;j++) 38 dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]); 39 for(j=1;j<=m;j++) 40 for(k=1;k<=j;k++) 41 dp[i][j][k]=dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]); 42 s[i][0]=dp[i][m][0]; 43 for(j=1;j<=m;j++) 44 s[i][j]=s[i][j-1]+dp[i][m][j]; 45 } 46 p1=p2=1.0; 47 for(i=1;i<=t;i++) 48 p1*=(s[i][m]-s[i][0]); 49 for(i=1;i<=t;i++) 50 p2*=(s[i][n-1]-s[i][0]); 51 printf("%.3f\n",p1-p2); 52 53 } 54 return 0; 55 }
