HDU 4576（概率DP+滚动数组）

刚开始用搜索写着玩了玩，超时是必然，后来看题解可以才知道搜索到的当前深度所有点的可能性（即概率）是本层搜索中所有到达本点概率的和。所以用两个数组分别存上一次的概率和更新后的概率就好了。原来概率DP就是求概率，然后用动态规划推导，滚动数组就是不断的刷新一个数组，一个交换位置用，算是见识了。加油加油！
Description
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. 
Input
There are multiple test cases. Each test case contains several lines. The first line contains four integers: above mentioned n(1≤n≤200) ,m(0≤m≤1,000,000),l,r(1≤l≤r≤n). Then m lines follow, each representing a command. A command is a integer w(1≤w≤100) representing the cell length the robot will walk for this command.   The input end with n=0,m=0,l=0,r=0. You should not process this test case. 
Output
For each test case in the input, you should output a line with the expected possibility. Output should be round to 4 digits after decimal points.
Sample Input
3 1 1 215 2 4 4120 0 0 0
Sample Output
0.50000.2500

#include <bits/stdc++.h>using namespace std;typedef pair<int, int> P;typedef long long LL;#define INF 0x3f3f3f3f#define PI acos(-1)#define MAX_N 10000//#define LOCALint n, m, l, r;double dp[2][205];int main(){#ifdef LOCALfreopen("b:\\data.in.txt", "r", stdin);#endif    while(scanf("%d%d%d%d", &n, &m, &l, &r))    {        if(!n && !m && !l && !r)            break;        memset(dp, 0, sizeof(dp));        dp[0][0] = 1;        int t = 0, k;//滚动数组        for(int i = 0; i < m; i++)        {            int mov;            scanf("%d", &mov);            k = t^1;            for(int c = 0; c < n; c++)                dp[k][c] = 0;            for(int j = 0; j < n; j++)            {                if(!dp[t][j])                    continue;                dp[k][(j+mov)%n] += dp[t][j]*0.5;                dp[k][(j-mov+n)%n] += dp[t][j]*0.5;            }            t = k;        }//        for(int i = 0; i < n; i++)//            cout << dp[t][i] << " " ;//        cout << endl;        double ans = 0;        for(int i = l; i <= r; i++)            ans += dp[t][i-1];        printf("%.4lf\n", ans);    }    return 0;}