poj2151 Check the difficulty of problems

poj2151 Check the difficulty of problems

​​ 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

2 2 2 0.9 0.9 1 0.9 0 0 0 Sample Output

0.972 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..

#include#include#includeusing namespace std;int n,m,t;double dp[1100][33][33],s[1100][33],p[1100][33];int main(){ freopen("poj2151.in","r",stdin); while(1){ scanf("%d%d%d",&m,&t,&n); if (!m&&!t&&!n) return 0; memset(dp,0,sizeof(dp)); 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){ dp[i][0][0]=1; for (int j=1;j<=m;++j){ for (int k=j-1;~k;--k){ dp[i][j][k+1]+=dp[i][j-1][k]*p[i][j]; dp[i][j][k]=dp[i][j-1][k]*(1-p[i][j]); } }s[i][0]=dp[i][m][0]; for (int j=1;j<=m;++j) s[i][j]=s[i][j-1]+dp[i][m][j]; }double p1=1,p2=1; for (int i=1;i<=t;++i) p1*=(1-s[i][0]),p2*=(s[i][n-1]-s[i][0]); printf("%.3f\n",p1-p2); } return 0;}

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