【HDU4576】Robot

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 2

15 2 4 4120 0 0 0

Sample Output

0.50000.2500

题解

直接dp+滚动数组。

采用刷表法，dp[i][j]表示第i个操作到第j个数字的概率，用dp[i][j]更新dp[i+1][j+w]和dp[j+1][j-w],注意事项都在代码中标注。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=210;double f[2][maxn];int main(){int n,m,l,r,w,now,last;while(scanf("%d%d%d%d",&n,&m,&l,&r)!=EOF){if(n==0&&m==0&&l==0&&r==0) break;//可能不是有0就停止，多次读入时注意 memset(f,0,sizeof(f));f[0][0]=1;now=1; while(m--){scanf("%d",&w);w%=n;//可能一开始大于n for(int j=0;j<n;j++)f[now][j]=0;for(int j=0;j<n;j++){//因为要转圈取模，所以只能从0开始符合规则，切记！！！ if(!f[now^1][j]) continue;f[now][(w+j)%n]+=0.5*f[now^1][j]; f[now][(j-w+n)%n]+=0.5*f[now^1][j];}now^=1;}double ans=0;for(int i=l;i<=r;i++)ans+=f[now^1][i-1];printf("%.4lf\n",ans);}       return 0;}