題目 color ger blog limit pos fine rtai oid

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4403		Accepted: 1941

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

/*題意：
ACM比賽中，共M道題，T個隊，pij表示第i隊解出第j題的概率
問 每隊至少解出一題且冠軍隊至少解出N道題的概率。


分析：
對於須要求得概率比較easy想到：
如果p1為每一個隊至少解出一題的概率，這個easy算出。
如果p2為每一個隊至少解出一題可是不超過n-1題的概率
所以終於答案為：p1-p2
如今問題是怎樣求出p2?
 

如果dp[i][j]表示第i個隊解出的題目<=j的概率
則dp[i][j]=解出1題+解出2題+...解出j題的概率
如今問題轉化為怎樣求解出1。2，3...k題的概率
如果x[i][j][k]表示第i個隊在前j題解出k題的概率
則:
x[i][j][k]=x[i][j-1][k-1]*p[i][j]+x[i][j-1][k]*(1-p[i][j]);
所以x[i][M][k]表示的就是第i個隊解出k題的概率 
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=1000+10;
const int N=30+10;
int m,t,n,index;
double x[2][N],dp,p[MAX][N];

int main(){
	while(cin>>m>>t>>n,m+t+n){
		double p1=1,p2=1,temp=1;
		for(int i=1;i<=t;++i){
			temp=1;
			for(int j=1;j<=m;++j){cin>>p[i][j];temp=temp*(1-p[i][j]);}
			p1=p1*(1-temp);//1-temp表示至少解出一道題 
		}
		for(int i=1;i<=t;++i){
			index=0;
			memset(x,0,sizeof x);//初始化前0個解出1~m為0 
			x[0][0]=1;//前0個解出0個為1 
			for(int j=1;j<=m;++j){
				index=index^1;
				for(int k=1;k<=m;++k){
					x[index][k]=p[i][j]*x[index^1][k-1]+(1-p[i][j])*x[index^1][k];
				}
				x[index][0]=x[index^1][0]*(1-p[i][j]);//表示前j道題做出0題的概率 
			}
			dp=0;//dp表示第i隊解出題目為1~n-1的概率 
			for(int j=1;j<=n-1;++j)dp+=x[index][j];
			p2=p2*dp;//p2表示解出1~n-1題的概率 
		}
		printf("%.3f\n",p1-p2);
	}
	return 0;
}

poj2151之概率DP