poj 2151 Check the difficulty of problems （检查问题的难度）

 1 /*
 2 Problem: 2151        User: bibier
 3 Memory: 5120K        Time: 47MS
 4 Language: G++        Result: Accepted
 5 */
 6 
 7 #include <stdio.h>
 8 #include <string.h>
 9 #include <iostream>
10 #include <algorithm>
11 #include <cstdio>
12 #include <cstring>
13 #include <cmath>
14 #include <stack>
15 #include <queue>
16 #include <vector>
17 using namespace std;
18 #define M 0x0f0f0f0f
19 #define min(a,b) (a>b?b:a)
20 #define max(a,b) (a>b?a:b)
21 
22 
23 float f[1003][33];//f[i][j]第i队做对第j个题的概率
24 float g[1003][33][33];//g[i][j][k]第i队前j个题做对k个题的概率
25 
26 //g[i][j][k] = g[i][j-1][k-1]*f[i][j] + g[i][j-1][k]*(1-f[i][j])
27 int main()
28 {
29     int tn,dn,n;
30     int i,j,k;
31     while(scanf("%d %d %d",&tn,&dn,&n)!=EOF)
32     {
33         if(tn==0&&dn==0&&n==0)
34             break;
35         memset(g,0,sizeof(g));
36         memset(f,0,sizeof(f));
37 
38         for(i=1; i<=dn; i++)
39             for(j=1; j<=tn; j++)
40                 scanf("%f",&f[i][j]);
41 
42         for(i=1; i<=dn; i++)
43         {
44             g[i][1][0]=1-f[i][1];
45             g[i][1][1]=f[i][1];
46         }
47 
48         for(i=1; i<=dn; i++)
49             for(j=2; j<=tn; j++)
50                 for(k=0; k<=j; k++)
51                 {
52                     if(k==j)
53                         g[i][j][k] = g[i][j-1][k-1]*f[i][j];
54                     else if(k==0)
55                         g[i][j][k] = g[i][j-1][k]*(1-f[i][j]);
56                     else
57                         g[i][j][k] = g[i][j-1][k-1]*f[i][j] + g[i][j-1][k]*(1-f[i][j]);
58                 }
59         float all=1;
60         for(i=1; i<=dn; i++) //每队至少做对一道题的最终概率
61             all*=(1-g[i][tn][0]);
62 
63         float midall=1;
64         for(i=1; i<=dn; i++)
65         {
66             float everyd=0;
67             for(j=1; j<n; j++)
68             {
69                 everyd+=g[i][tn][j];
70             }
71             midall*=everyd;
72         }
73         all-=midall;
74         printf("%.3f\n",all);
75     }
76     return 0;
77 }