D - Maximum Random Walk
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice HDU
4487
Appoint description:
System Crawler (2016-05-03)
Description
Consider the classic random walk: at each step, you have a 1/2 chance of taking a step to the left and a 1/2 chance of taking a step to the right. Your expected position after a period of time is zero; that is, the average over many
such random walks is that you end up where you started. A more interesting question is what is the expected rightmost position you will attain during the walk.
Input
The first line of input contains a single integer P, (1 <= P <= 15), which is the number of data sets that follow. Each data set should be processed identically and independently.
Each data set consists of a single line of input consisting of four space-separated values. The first value is an integer K, which is the data set number. Next is an integer n, which is the number of steps to take (1 <= n <= 100). The final two are double precision
floating-point values L and R
which are the probabilities of taking a step left or right respectively at each step (0 <= L <= 1, 0 <= R <= 1, 0 <= L+R <= 1). Note: the probably of not taking a step would be 1-L-R.
Output
For each data set there is a single line of output. It contains the data set number, followed by a single space which is then followed by the expected (average) rightmost position you will obtain during the walk, as a double precision
floating point value to four decimal places.
Sample Input
3 1 1 0.5 0.5 2 4 0.5 0.5 3 10 0.5 0.4
Sample Output
1 0.5000 2 1.1875 3 1.4965
一个人在原点随机走n步 每次向左走的概率为L，向右走的概率为R 不走的概率是 1-R-L
问走的过程中向右走最远的期望
设dp【i，j，k】表示第i步走到j了此时曾经到过最远距离为k的概率
ACcode：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
double dp[2][202][202];
int main(){
int loop;
scanf("%d",&loop);
while(loop--){
int cnt,step,k;
double l,r,s,ans;
scanf("%d%d%lf%lf",&cnt,&step,&l,&r);
s=1.0-l-r;
memset(dp,0,sizeof(dp));
dp[0][100][100]=1;
for(int i=1;i<=step;++i)
for(int j=100-i;j<=100+i;j++){
for(k=j>100?j:100;k<=100+i;++k){
if(j==k)
dp[i&1][j][k]=dp[(i+1)&1][j][k]*s+dp[(i+1)&1][j-1][k-1]*r+dp[(i+1)&1][j-1][k]*r;
else
dp[i&1][j][k]=dp[(i+1)&1][j][k]*s+dp[(i+1)&1][j-1][k]*r+dp[(i+1)&1][j+1][k]*l;
}
}
ans=0;
for(int j=101;j<=101+step;++j)for(int i=100-step;i<=100+step;++i)ans+=dp[step&1][i][j]*(j-100);
printf("%d %.4f\n",cnt,ans);
}
return 0;
}