hdu 4576 概率dp **

题意:Michael has a telecontrol robot. One day he put the robot on a loop with n cells. The cells are numbered from 1 to n clockwise.

At
first the robot is in cell 1. Then Michael uses a remote control to
send m commands to the robot. A command will make the robot walk some
distance. Unfortunately the direction part on the remote control is
broken, so for every command the robot will chose a direction(clockwise
or anticlockwise) randomly with equal possibility, and then walk w cells
forward.
Michael wants to know the possibility of the robot stopping in the cell that cell number >= l and <= r after m commands.

链接:点我

时间上卡的有点紧

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>
 #include<map>
 using namespace std;
 #define MOD 1000000007
 const int INF=0x3f3f3f3f;
 const double eps=1e-;
 typedef long long ll;
 #define cl(a) memset(a,0,sizeof(a))
 #define ts printf("*****\n");
 const int MAXN=;
 int n,m,tt;
 double dp[][MAXN];
 int a[MAXN];
 int main()
 {
     int i,j,k;
     #ifndef ONLINE_JUDGE
     freopen("1.in","r",stdin);
     #endif
     int x,l,r;
     while(scanf("%d%d%d%d",&n,&m,&l,&r)!=EOF)
     {
         if(n==&&m==&&l==&&r==)break;
         dp[][]=;
         for(i=;i<=n;i++)    dp[][i]=;
         int now=;
         while(m--)
         {
             scanf("%d",&x);
             for(i=;i<n;i++)
             {
                 dp[now^][i]=;
             }
             for(i=;i<n;i++)
             {
                 if(dp[now][i]==)   continue;
                 dp[now^][((i-x)%n+n)%n]+=0.5*dp[now][i];
                 dp[now^][(i+x)%n]+=0.5*dp[now][i];
             }
             now^=;
         }
         double ans=;
         for(i=l-;i<r;i++)
         {
             ans+=dp[now][i];
         }
         printf("%.4lf\n",ans);
     }
 }