Check the difficulty of problems
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 3617 | Accepted: 1580 |
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------------------------------------------------------
f[i][j][k]表示第i组队伍做前j道题有k道正确的概率。。。
显然有 f[i][j][k]=f[i][j-1][k]*(1-p[i][j])+f[i][j-1][k-1]*p[i][j];
至少有一组做出n道以上的题且每组至少做出1道题的概率等于每组至少做出1题的概率-没有一组做出n道题以上的概率。
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#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn=1111; int m,t,n; double p[maxn][44]; double f[maxn][44][44]; int main() { while (~scanf("%d%d%d",&m,&t,&n)) { if (m==0&&t==0&&n==0) break; memset(f,0,sizeof(f)); memset(p,0,sizeof(p)); for (int i=1;i<=t;i++) { for (int j=1;j<=m;j++) { scanf("%lf",&p[i][j]); } } for (int i=1;i<=t;i++) { f[i][0][0]=1; for (int j=1;j<=m;j++) { for (int k=0;k<=j;k++) { if (k==0) { f[i][j][k]=f[i][j-1][k]*(1-p[i][j]); } else { f[i][j][k]=f[i][j-1][k]*(1-p[i][j])+f[i][j-1][k-1]*p[i][j]; } } } } double ans=1,sum=0,ans2=1; for (int i=1;i<=t;i++) { sum=0; for (int j=1;j<=m;j++) { sum+=f[i][m][j]; } ans*=sum; } for (int i=1;i<=t;i++) { sum=0; for (int j=1;j<n;j++) { sum+=f[i][m][j]; } ans2*=sum; } ans-=ans2; printf("%0.3lf\n",ans); } return 0; }