POJ2151Check the difficulty of problems【概率dp求概率】
2015-08-24 18:00
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Check the difficulty of problems
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 |
#include<cstdio> #include<cstdlib> #include<cstring> using namespace std; double pr[1010][35][35];// double dp[1010][35]; double p[1010][35]; int main() { int M,T,N,i,j,k; double p1,pN; while(scanf("%d%d%d",&M,&T,&N),M||T||N){ memset(dp,0,sizeof(dp)); for(i=1;i<=T;++i){ for(j=1;j<=M;++j){ scanf("%lf",&p[i][j]); } } memset(pr,0,sizeof(pr)); memset(dp,0,sizeof(dp)); p1=pN=1.0; for(i=1;i<=T;++i){ pr[i][0][0]=1; for(j=1;j<=M;++j){ pr[i][j][0]=pr[i][j-1][0]*(1.0-p[i][j]); } for(j=1;j<=M;++j){ for(k=1;k<=M;++k){ pr[i][j][k]=pr[i][j-1][k-1]*p[i][j]+pr[i][j-1][k]*(1-p[i][j]);//前j道题中做出k道题的概率 } } dp[i][0]=pr[i][M][0]; for(j=1;j<=M;++j){ dp[i][j]=dp[i][j-1]+pr[i][M][j];//做出j道题的概率 } p1=p1*(dp[i][M]-dp[i][0]);//解决题数在1~M之间的概率 pN=pN*(dp[i][N-1]-dp[i][0]);//解决题数在1~N-1之间的概率 即没有队做出N道题以上的概率 } printf("%.3lf\n",p1-pN); } return 0; }
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