hdu4576Robot
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Robot
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)Total Submission(s): 1783 Accepted Submission(s): 655
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.
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 1 215 2 4 4120 0 0 0
Sample Output
0.50000.2500
Source
2013ACM-ICPC杭州赛区全国邀请赛
做这道题时我很无语,因为是挑一道最简单的题来做,但是自己真的很菜,很水!!
明知道这道题是水题都是在看到人家的结题报告才知道这是概率dp。
对自己的水平很失望。。
还有我看到dp的循环次数时以为会超时(一百万*2*200)这样都不超时,Orz~~Orz~~Orz~~Orz~~Orz!!
用b数组代表是这个状态,a数组代表是上一个状态 然后有
b[(j-w+n)%n]+= a[j]*0.5;
b[(j+w)%n]+= a[j]*0.5;
#include<cstdio>#include<cstring>double a[212],b[212];int main(){ int n,m,l,r,w,j; double va; while(scanf("%d%d%d%d",&n,&m,&l,&r)) { if(n+m+l+r==0) break; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); a[0] = 1.0; for(int i=0;i<m;i++) { scanf("%d",&w); for(j=0;j<n;j++) { if(a[j]>0)//记住一定要这个剪枝,不然就很容易超时 { va = a[j]*0.5; b[(j-w+n)%n]+= va; b[(j+w)%n]+= va; } } for(j=0;j<n;j++) { a[j] = b[j]; b[j] = 0; //还有救市这个一定要初始化为0,不然就会错。 } } l--; r--; double ans = 0.0; for(;l<=r;l++) ans+=a[l]; printf("%0.4f\n",ans); } return 0;}
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