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-5;
typedef long long ll;
#define cl(a) memset(a,0,sizeof(a))
#define ts printf("*****
");
const int MAXN=220;
int n,m,tt;
double dp[2][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==0&&m==0&&l==0&&r==0)break;
        dp[0][0]=1;
        for(i=1;i<=n;i++)    dp[0][i]=0;
        int now=0;
        while(m--)
        {
            scanf("%d",&x);
            for(i=0;i<n;i++)
            {
                dp[now^1][i]=0;
            }
            for(i=0;i<n;i++)
            {
                if(dp[now][i]==0)   continue;
                dp[now^1][((i-x)%n+n)%n]+=0.5*dp[now][i];
                dp[now^1][(i+x)%n]+=0.5*dp[now][i];
            }
            now^=1;
        }
        double ans=0;
        for(i=l-1;i<r;i++)
        {
            ans+=dp[now][i];
        }
        printf("%.4lf
",ans);
    }
}