Maximum Random Walk

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 219    Accepted Submission(s): 120

Problem 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
Source 思路： 用三維陣列來表示狀態：dp[i][j][k]  i-走了i步 j-到達第j個位置 k-右移最大距離為k，遞推完後求期望。 程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 201
using namespace std;

int n,m;
float pl,pr,pm,ans;
float dp[101][maxn][maxn];

int main()
{
    int i,j,k,t,test,l,r;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%f%f",&test,&n,&pl,&pr);
        pm=1-pl-pr;
        memset(dp,0,sizeof(dp));
        dp[0][100][100]=1;            // 將一百設為起點
        l=100-n;
        r=100+n;
        for(i=1;i<=n;i++)             // 走了i步
        {
            for(j=l;j<=r;j++)         // 到達第j個位置
            {
                for(k=100;k<=r;k++)   // 最大到達的右邊界
                {
                    dp[i][j][k]+=dp[i-1][j][k]*pm;
                    dp[i][j-1][k]+=dp[i-1][j][k]*pl;
                    if(j+1>k) dp[i][j+1][j+1]+=dp[i-1][j][k]*pr;
                    else dp[i][j+1][k]+=dp[i-1][j][k]*pr;
                }
            }
        }
        ans=0;
        for(k=100;k<=r;k++)
        {
            for(j=l;j<=r;j++)
            {
                ans+=dp[n][j][k]*(k-100);
            }
        }
        printf("%d %.4f\n",test,ans);
    }
    return 0;
}