POJ 2151 Check the difficulty of problems （概率DP）

POJ 2151 Check the difficulty of problems （概率DP）

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: All of the teams solve at least one problem.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 20.9 0.91 0.90 0 0

Sample Output

0.972

题意

ACM比赛中，共 M 道题，T 个队， p[i][j] 表示第 i 队解出第 j

思路

首先，每队至少解出一题中一定包含所有队解题数目至少为1并且都是少于 n 的，如果减去这一部分便是至少有一个队伍的解题数目大于 n 了，至于冠军队嘛！我们不用关心，因为我们只需要让它大于 n

ans: 所有队至少解出一题的概率as : 所有队至少解出一题但是不超过 n−1 题的概率最终答案： ans−as一个队伍解出不超过 n−1 道题目的概率是解出 1、2、...、n−1pd[j][k] 表示在前 j 道题中做对 kpd[j][k]=pd[j−1][k]∗(1.0−p[i][j])+pd[j−1][k−1]∗p[i][j]其中考虑 j$j$ 题，要么做对，要么做错，只有这两种情况，其概率相加。

AC 代码

#include#include#include#include#include#includeusing namespace std;#include#includedouble pd[35][35],p[1005][35]; //pd[i][j]表示在前i道题中做对j道的概率int main(){ int m,t,n; while(~scanf("%d%d%d",&m,&t,&n)&&(m||t||n)) { double ans=1.0,as=1.0; //ans 为所有队至少A一道题的概率 as为所有队A的题目数量都不超过n的概率 memset(pd,0,sizeof(pd)); for(int i=0; i

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