Xdoj 1020 Acmer to brush the problem (basic DP)

Topic Description Brush problem is each acmer way, known to have N on a OJ topic, the first topic small x can do the probability for Pi (0<=pi<=1,1<=i<=n) small x to do at least the probability of the K-Question input the first line input a positive integer t, (t< =20), indicating that there is a T-group test Sample. The second line enters a positive integer n,k, (1<=n,k<=1000) the third line inputs n decimals, respectively, Pi (1<=i<=n,0<=pi<=1), indicating the probability that small x does the first topic. Output small x do at least the probability of a K-question, and wrap (keep 4 decimal) sample input

21 10.53 20.3) 0.2 0.1

Sample output

0.50000.0980

Tips

Main topic:

This is to give you a probability of doing the right thing for each question, and then, let you find out at least how much the probability of doing the K-question.

Problem Solving Ideas:

In fact, it is not difficult, as long as the launch to Dp[i][j] play just fine, a start to push for a while, and less a condition (q God to the nest to fill up)

State: Dp[i][j] represents the probability of having an I-question, which is a problem for J-questions.

Initial state: Dp[0][0] = 1,dp[0][1] = 0;

State transition equation: dp[i][j] = dp[i-1][j-1]*p[i]+dp[i-1][j]* (1-p[i]);

Dp[i][0] = (1-p[i]) *dp[i-1][0]

Finally only asked to come out sum+=dp[n][k]+dp[n][k+1]+dp[n][k+2]+...+dp[n][n];

Code:

1# include<cstdio>2# include<iostream>3  4 using namespacestd;5  6# define MAX12347  8 DoubleP[max];9 DoubleDp[max][max];Ten intn,k; One   A intMainvoid) - { -     intT;SCANF ("%d",&t); the      while(t-- ) -     { -scanf"%d%d",&n,&k); -          for(inti =1; I <= n;i++ ) +         { -scanf"%LF",&p[i]); +         } Adp[0][0] =1; atdp[0][1] =0; -          for(inti =1; I <= n;i++ ) -         { -dp[i][0] = (1-p[i]) *dp[i-1][0]; -         } -          for(inti =1; I <= n;i++ ) in         { -              for(intj =1; J <= n;j++ ) to             { +DP[I][J] = (1-p[i]) *dp[i-1][j]+p[i]*dp[i-1][j-1]; -             } the         } *         Doublesum =0; $          for(inti = K;i <= n;i++ )Panax Notoginseng         { -sum+=Dp[n][i]; the         } +printf"%.4lf\n", sum); A   the     } +   -   $     return 0; $}

Xdoj 1020 Acmer to brush the problem (basic DP)