题意：在acm比赛中，n题，t队。给出每个队做对每题的概率，问每队至少对一题，，至少有一队做对至少m题的概率分析：dp，f[i][j]表示第i个队伍做对第j题的概率。g[i][j][k]表示第i个队伍对于前j题而言做对k道的概率。g[i][j][k] = g[i][j – 1][k – 1] * (f[i][j]) + g[i][j – 1][k] * (1 – f[i][j]);有了所有的g，我们就可以求出每个队至少做对1题的概率：ans *= 1 – g[i][n][0];
再减去每个队都只做对1～m-1题的概率(把每个队做对1～m-1题的概率加和，并把各队结果相乘）
Time Limit:2000MSMemory Limit:65536K
Total Submissions:5375Accepted:2371
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
#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;double f[1001][31];double g[1001][31][31];int main(){int n,t,m;while(~scanf("%d %d %d",&n,&t,&m)){if(n+t+m==0)break;for(int i=1; i<=t; i++)for(int j=1; j<=n; j++)scanf("%lf",&f[i][j]);memset(g,0,sizeof(g));for(int i=1; i<=t; i++){g[i][0][0]=1;for(int j=1; j<=n; j++){g[i][j][0]=g[i][j-1][0]*(1-f[i][j]);for(int k=1; k<=j; k++){g[i][j][k] = g[i][j -1][k -1] * (f[i][j])+ g[i][j -1][k] * (1- f[i][j]);}}}double ans=1;for(int i=1; i<=t; i++)ans*=1-g[i][n][0];//ans=1-ans;double ans2=1;for(int i=1; i<=t; i++){double ans1=0;for(int j=1; j<m; j++)ans1+=g[i][n][j];ans2*=ans1;}ans-=ans2;printf("%.3f\n",ans);}return 0;}不算太难的一道题目，但是错了好几遍原因是poj的%lf要换成%f真心给跪了、。。。。
贪婪是最真实的贫穷，满足是最真实的财富
