Check the difficulty of problems
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 4011 | Accepted: 1779 |
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
Source
POJ Monthly,鲁小石
设a[i][j][k]表示第i队在前j道题中共解出k道题的概率,易得a[i][j][k]有如下递推
关系(另需考虑边界条件):
a[i][j][k] = a[i][j-1][k-1] * p[i][j] + a[i][j-1][k] * (1-p[i][j])
设s[i][j]表示a[i][M][0] + a[i][M][1] + ... + a[i][M][j]
问题的解可以转化为:每队均至少做一题的概率(用P1表示)减去每队做题数均在1到N-1
之间的概率(用P2表示)。
P1 = (s[1][M] - s[1][0])*(s[2][M]-s[2][0])*...*(s[T][M]-s[T][0])
P2 = (s[1][N-1] - s[1][0])*(s[2][N-1]-s[2][0])*...*(s[T][N-1]-s[T][0])
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 5 using namespace std; 6 7 double dp[1100][40][40]; 8 double p[1100][40]; 9 double sum[1100][40]; 10 double p1,pn; 11 int M,T,N; 12 13 int main() 14 { 15 while(scanf("%d%d%d",&M,&T,&N)!=EOF) 16 { 17 if((M||T||N)==0) break; 18 19 memset(dp,0,sizeof(dp)); 20 memset(sum,0,sizeof(sum)); 21 memset(p,0,sizeof(p)); 22 23 for(int i=1;i<=T;i++) 24 { 25 p[i][0]=1; 26 for(int j=1;j<=M;j++) 27 { 28 scanf("%lf",&p[i][j]); 29 } 30 } 31 32 for(int i=1;i<=T;i++) 33 { 34 dp[i][0][0]=1; 35 } 36 37 for(int i=1;i<=T;i++) 38 { 39 for(int j=1;j<=M;j++) 40 { 41 for(int k=0;k<=j;k++) 42 { 43 dp[i][j][k]=dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]); 44 } 45 } 46 } 47 48 for(int i=1;i<=T;i++) 49 { 50 for(int j=0;j<=M;j++) 51 { 52 for(int k=0;k<=j;k++) 53 sum[i][j]+=dp[i][M][k]; 54 } 55 } 56 /* 57 for(int i=1;i<=T;putchar(10),i++) 58 { 59 for(int j=1;j<=M;putchar(10),j++) 60 { 61 for(int k=0;k<=M;k++) 62 { 63 printf("dp[%d][%d][%d]: %lf ",i,j,k,dp[i][j][k]); 64 } 65 } 66 } 67 68 for(int i=1;i<=T;putchar(10),i++) 69 { 70 for(int k=0;k<=M;k++) 71 { 72 printf("sum[%d][%d]: %lf ",i,k,sum[i][k]); 73 } 74 } 75 */ 76 p1=1;pn=1; 77 for(int i=1;i<=T;i++) 78 { 79 p1*=(sum[i][M]-sum[i][0]); 80 pn*=(sum[i][N-1]-sum[i][0]); 81 } 82 printf("%.3lf ",p1-pn); 83 } 84 85 return 0; 86 }