Robot
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 5363 Accepted Submission(s): 1584
Problem 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.
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
1
5 2 4 4
1
2
2
0 0 0 0
Sample Output
0.5000
0.2500
Source
题意：
有长度为n的环，有m次操作，每次输入一个数w，可以选择顺时针或者逆时针走w步，问最后停在l区间[l,r]的概率。最初在1点。
输入n，m，l，r；
代码：
//这一步的状态只有上一步的状态决定，可列转移方程，但显然要用滚动数组优化。
//注意的是中间过程中有太多的取模运算，直接算会超时所以先预处理出来取模的值再算
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m,l,r,mp[];
double dp[][];
int main()
{
while(scanf("%d%d%d%d",&n,&m,&l,&r)==){
if(n==&&m==&&l==&&r==) break;
for(int i=;i<=;i++) mp[i]=i%n;
int x;
l--;r--;
memset(dp,,sizeof(dp));
dp[][]=;
for(int i=;i<=m;i++){
int I=i&;
scanf("%d",&x);
x%=n;
for(int j=;j<n;j++){
int tmp=mp[j+x];
dp[I][tmp]+=dp[!I][j]*0.5;
tmp=mp[j+(n-x)];
dp[I][tmp]+=dp[!I][j]*0.5;
}
memset(dp[!I],,sizeof(dp[!I]));
}
double sum=;
int I=m&;
for(int i=l;i<=r;i++)
sum+=dp[I][i];
printf("%.4lf\n",sum);
}
return ;
}