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?
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
比较简单的一道概率dp,然而还是没能自己想出来(现在一丢丢罪恶感都没。。。)
要求的概率=P(所有队伍都不挂机)-P(所有队伍的过题数都在[1,n-1]之间)
在DP的时候开一个三维数组:dp[i][j][k]表示第i支队伍在前j题中通过了k道题,这样在最后可以看dp[i][m][k]来得到每个队伍过某题数的概率。递推方程很好得出:dp[i][j][k]=dp[i][j-1][k]*(1-p[i][j])+dp[i][j-1][k-1]*p[i][j];
注意:①k=0时是特殊情况,dp值要单独计算!
②提交时看好自己的printf格式,如果是G++需要将double型用%.3f输出,C++用%lf输出。
#include<iostream> #include<cstdio> #include<vector> #include<set> #include<map> #include<string.h> #include<cmath> #include<algorithm> #include<queue> #include<stack> #define LL long long #define mod 1000000007 #define inf 0x3f3f3f3f using namespace std; double a[1100][55]; double dp[1100][55][55]; int main() { int m,t,n; while(~scanf("%d%d%d",&m,&t,&n)) { if(t==0&&m==0&&n==0) break; memset(a,0,sizeof(a)); memset(dp,0,sizeof(dp)); for(int i=1;i<=t;i++) for(int j=1;j<=m;j++) scanf("%lf",&a[i][j]); for(int i=1;i<=t;i++) { dp[i][1][0]=1-a[i][1]; dp[i][1][1]=a[i][1]; } for(int i=1;i<=t;i++) for(int j=2;j<=m;j++) for(int k=0;k<=j;k++) { if(k==0) dp[i][j][k]=dp[i][j-1][k]*(1-a[i][j]); else dp[i][j][k]=dp[i][j-1][k-1]*a[i][j]+dp[i][j-1][k]*(1-a[i][j]); } double sum=1; for(int i=1;i<=t;i++) sum*=(1-dp[i][m][0]); double ans=1; for(int i=1;i<=t;i++) { double tem=0; for(int j=1;j<=n-1;j++) tem+=dp[i][m][j]; ans*=tem; } printf("%.3lf ",sum-ans); } return 0; }