有 T(1<T<=1000) 支队伍和 M(0<M<=30) 个题目，已知每支队伍 i 解决每道题目 j 的的概率 p[i][j]，现在问：每支队伍至少解决一道题，且解题最多的队伍的题目数量不少于 N(0<N<=M) 的概率是多少？
p[i][j]表示第i个队伍做对第j题的概率。dp[i][j][k]表示第i个队伍在前j题中做对了k道的概率。
dp[i][j][k] = dp[i][j-1][k-1]*(p[i][j])+dp[i][j-1][k]*(1-p[i][j]);
再求出每个队都至少做对 1 道题的概率：ans1 *= 1 - dp[i][m][0];
求出每个队都只做对了 1 ～ n-1 题的概率 ans2即：(把每个队做对 1 ～ n-1 题的概率相加后，并把每个队的结果相乘）；
然后两者相减ans1-ans2
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4 int M,T,N;
5 double p[1002][32];
6 double dp[1002][32][32];
7 double sum[1002][32];
8 int main() {
9 scanf("%d %d %d",&M,&T,&N);
10 while(M&&N&&T){
11 memset(p,0,sizeof(p));
12 for(int i=1;i<=T;i++){
13 for(int j=1;j<=M;j++){
14 scanf("%lf",&p[i][j]);
15 }
16 }
17 memset(dp,0,sizeof(dp));
18 for(int i=1;i<=T;i++){
19 dp[i][1][1]=p[i][1];
20 dp[i][1][0]=1-p[i][1];
21
22 }
23 for(int i=1;i<=T;i++){
24 for(int j=2;j<=M;j++){
25 for(int k=0;k<=j;k++){
26 dp[i][j][k]=dp[i][j-1][k]*(1.0-p[i][j]);
27 if(k>0) dp[i][j][k]+=dp[i][j-1][k-1]*p[i][j];
28 }
29 }
30 }
31 memset(sum,0,sizeof(sum));
32 for(int i=1;i<=T;i++){
33 sum[i][0]=dp[i][M][0];
34 for (int j=1;j<=M;j++) {
35 sum[i][j]=sum[i][j-1]+dp[i][M][j];
36 }
37 }
38 double ans1=1.0,ans2=1.0;
39 for(int i=1; i<=T; i++) {
40 ans1*=sum[i][M]-sum[i][0];
41 ans2*=(sum[i][N-1]-sum[i][0]);
42 }
43 printf("%.3lf\n",ans1-ans2);
44 scanf("%d %d %d",&M,&T,&N);
45 }
46
47 return 0;
48 }
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 5766 | Accepted: 2515 |
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