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 poj 2151 概率dp

发布时间: 2015-02-14 02:12


直播性爱视频//poj 2151 概率dp

 1 #include "iostream"
 2 #include "cstdio"
 3 #include "cstring"
 4 #include "algorithm"
 5 using namespace std;
 6 double dp[33][33]; 
 7 int M, T, N;    //problem, team, least
 8 double p[1010][33];
 9 int main()
10 {
11     int i, j, k;
12     while(scanf("%d%d%d", &M, &T, &N) && (M || T || N)) {
13         for(i = 1; i <= T; ++i)
14             for(j = 1; j <= M; ++j)
15                 scanf("%lf", &p[i][j]);
16         double res_1 = 1, res_2 = 1;    //所有队伍都解出一题以上的概率,所有队伍都没解到超过 N 道的概率
17         for(i = 1; i <= T; ++i) {
18             dp[0][0] = 1;
19             for(j = 1; j <= M; ++j) {
20                 for(k = 0; k <= j && k < N; ++k) {
21                     if(k >= 1)
22                         dp[j][k] = dp[j - 1][k] * (1 - p[i][j]) + dp[j - 1][k - 1] * p[i][j];
23                     else
24                         dp[j][k] = dp[j - 1][k] * (1 - p[i][j]);
25                 }
26             }
27             res_1 *= (1 - dp[M][0]);
28             double tmp = 0;
29             for(k = 1; k < N; ++k)
30                 tmp += dp[M][k];
31             res_2 *= tmp;
32         }
33         printf("%.3f\n",res_1 - res_2);
34     }
35 }

 

//优化下空间

 1 #include "iostream"
 2 #include "cstring"
 3 #include "algorithm"
 4 #include "cstdio"
 5 using namespace std;
 6 int M, T, N; //problem, team, least
 7 double dp[35][35];
 8 
 9 int main()
10 {
11     int i, j, k;
12     double p, res_1, res_2; //所有队伍都解出一题以上的概率,所有队伍都没解到超过 N 道的概率
13     while(scanf("%d%d%d", &M, &T, &N) && (N || T || M)) {
14         res_1 = res_2 = 1;
15         for(i = 1; i <= T; ++i) {
16             dp[0][0] = 1;
17             for(j = 1; j <= M; ++j) {
18                 scanf("%lf", &p);
19                 for(k = 0; k <= j && k < N; ++k) {
20                     if(k >= 1)
21                         dp[j][k] = dp[j - 1][k] * (1 - p) + dp[j - 1][k - 1] * p;
22                     else
23                         dp[j][k] = dp[j - 1][k] * (1 - p);
24                 }
25             }
26             res_1 *= (1 - dp[M][0]);
27             double tmp = 0;
28             for(k = 1; k<N; ++k)
29                 tmp += dp[M][k];
30             res_2 *= tmp;
31         }
32         printf("%.3f\n", res_1 - res_2);
33     }
34 }

 


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