Robot
Time Limit: 8000/4000 MS (Java/others) memory limit: 102400/102400 K (Java/Others)
Total submission (s): 2776 accepted submission (s): 948
Problem descriptionMichael 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.
InputThere 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 shoshould not process this test case.
OutputFor each test case in the input, you should output a line with the expected possibility. Output shocould be round to 4 digits after decimal points.
Sample input 3 1 215 2 4 4120 0 0 0
Sample output0.50000.2500
Source2013ACM-ICPC Hangzhou Division national invitational Competition
Problem solving: DP...
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 double dp[2][210];18 int main() {19 int n,m,l,r,tmp,i,cur;20 double ans = 0.0;21 while(scanf("%d %d %d %d",&n,&m,&l,&r),n||m||l||r){22 dp[0][0] = 1;23 for(int i = 1; i < n; i++) dp[0][i] = 0;24 cur = 0;25 while(m--){26 scanf("%d",&tmp);27 for(i = 0; i < n; i++)28 dp[cur^1][i] = 0.5*dp[cur][(i-tmp+n)%n] + 0.5*dp[cur][(i+tmp)%n];29 cur ^= 1;30 }31 ans = 0;32 for(i = l-1; i < r; i++)33 ans += dp[cur][i];34 printf("%.4f\n",ans);35 }36 return 0;37 }
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HDU 4576 Robot