POJ 2151 Check the difficulty of problems（概率dp）

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Check the difficulty of problems

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 5419 Accepted: 2384

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

POJ Monthly,鲁小石

题意: n个人，m道题，求每个人做出题，并且至少一个做出k道题的概率

思路:等于每个人做出题的概率--每个人做出题并且没有一个人做出k到题的概率

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b)  for(i = a; i < b; i++)#define free(i,b,a) for(i = b; i >= a;i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define ssf(n)      scanf("%s", n)#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define bug         pf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 1005double dp[N][35][35];  // dp[i][j][k]  第 i 个人 前 j 道题 做 k 道概率double p[N][35];int n,m,k;void solve(){int i,j,t;mem(dp,0);    fre(i,1,n+1)         //初始每一个人第一道题 对 与不对      {      dp[i][1][0]=1-p[i][1];dp[i][1][1]=p[i][1];      }fre(i,1,n+1)         //每一个人m到题都没有作对 { fre(j,2,m+1)  dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]); }fre(i,1,n+1)        //每一个人j道题 做t道概率 fre(j,2,m+1)  fre(t,1,j+1) {       if(t<j) dp[i][j][t]=dp[i][j-1][t]*(1-p[i][j])+dp[i][j-1][t-1]*p[i][j];   else     dp[i][j][t]=dp[i][j-1][t-1]*p[i][j]; } double p1=1,p2; fre(i,1,n+1)    p1*=(1-dp[i][m][0]);    //每个人都做出提的概率 p2=1; fre(i,1,n+1)  {   double te=0;   fre(j,1,k)     te+=dp[i][m][j]; p2*=te;            //每个人做出题但是没有一个做了k道题的概率  }pf("%.3lf\n",p1-p2);}int main(){int i,j;while(~sfff(m,n,k),n+m+k){fre(i,1,n+1)   fre(j,1,m+1)     scanf("%lf",&p[i][j]);   solve();}    return 0;}