Check the difficulty of problems
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 4522 | Accepted: 1993 |
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 20.9 0.91 0.90 0 0
Sample Output
0.972
Source
,鲁小石
题目大意：
有 M 道题目 T 支队伍。N表示 最好 的队 至少要做出N题 ，紧接下来T行M列，表示某队做出某题 的概率为p ，问你每支队至少做出1题，最好的队至少做出N题的概率是多少？
解题思路：
解题代码：一题动态规划的题。 既然最好的队至少做出N题。那么用二维记录，DP [t][f] 记录还剩 t 支队及是否出现超过N题的事件的概率。假设当前这支队伍做出超过N题，那么f置为1，否则还是f。弹了两遍，第一遍由于忘记算做出0题的情况，第二遍由于递归中数组开得略大些超内存了。
#include#include using namespace std;const int maxt=1100;const int maxn=32;double dp[maxt][2];double p[maxt][maxn];int T,M,N;double DP(int t,int f){ if(t<=0) return f; if(dp[t][f]>-1.0) return dp[t][f]; double a[maxn][maxn],ans=0; a[0][0]=1; for(int i=1;i<=M;i++){ for(int j=0;j<=i;j++){ a[i][j]=0; if(j-1>=0) a[i][j]+=a[i-1][j-1]*p[t][i]; if(i-1>=j) a[i][j]+=a[i-1][j]*(1-p[t][i]); } } for(int i=1;i
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