2015
09-16

# 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.

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.

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.

3 1 1 2
1
5 2 4 4
1
2
0 0 0 0

0.5000
0.2500

dp[(i + w) % n][j] = dp[i[j-1] * 0.5；

dp[i-w][j]=dp[i][j-1]*0.5;其中i-w不断的加n使它大于等于0；

dp[(i + w) % n][j&1] += dp[i][!(j&1)] * 0.5；
dp[x][j&1] += dp[i][!(j&1)] * 0.5；x=i-w，x也要叠加到非负；

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 220;
const double eps = 0.0;
double dp[N][2];

int main(){
int n, m, l, r;
while(cin >> n >> m >> l >> r){
if(n == 0 && m == 0 && l == 0 && r == 0)
break;
int w, i, j;
for(i = 0;i < n;i++)
dp[i][0] = dp[i][1] = eps;
dp[0][0] = 1.0;
for(j = 1;j <= m;j++){
scanf("%d", &w);
for(i = 0;i < n;i++){
dp[(i + w) % n][j&1] += dp[i][!(j&1)] * 0.5;
int x = i - w;
while(x < 0)
x += n;
dp[x][j&1] += dp[i][!(j&1)] * 0.5;
}
for(i = 0;i < n;i++)
dp[i][!(j&1)] = eps;
}
double res = 0.0;
l--, r--;
for(i = l;i <= r;i++)
res += dp[i][m&1];
printf("%.4lf\n", res);
}
return 0;
}