Maximum Random Walk
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 229 Accepted Submission(s): 128

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
Greater New York 2012

看了解题报告觉得挺简单的啊，自己写，写好了，怎么测试结果都是0，无奈把自己的代码和解题报告比对，对了好久都没发现问题，之后又比对了好多次，最后发现是我把输入向右走的概率输入那里写成了%f，天哪。。。就是这么个小错误，坑了我几个小时了，下次一定要小心再小心，坚决不能再犯这个错误了
题意：给你步数和向左走向右走的概率，剩下的是不走的概率，问你走过的最远距离(这里的意思不是说最后的距离，而是你经过的最远距离，不管是哪一步走的，只要是最远的就可以)的期望值。下面讲解我就copy一下啊
用dp[v][i][j]表示当前前位置在i。途中走到的最右边的位置为j。那么状态转移方程为：
往右走：
1.如果i+1>j。dp[v^1][i+1][i+1]+=dp[v][i][j]*r。
2.如果i+1<=j。dp[v^1][i+1][j]+=dp[v][i][j]*r。
往左走：
dp[v^1][i-1][j]+=dp[v][i][j]*l。

不走：
dp[v^1][i][j]+=dp[v][i][j]*(1-l-r)。
但需要注意的是。i可能为负数所以把起点往右移动n个位置(n为步数)。

#include   
#include   
double dp[2][205][105];  
int main()  
{  
    int i,j,k,n,t,no,c;  
    double x,y,z,ans;  
    scanf("%d",&t);  
    while(t--)  
    {  
        scanf("%d%d%lf%lf",&no,&n,&x,&y);  
        c=0;  
        z=1.0-x-y;  
        memset(dp[0],0,sizeof(dp[0]));  
        dp[0][n][0]=1;  
        for(k=0;k0)  
                    {  
                        if(i+1>j+n)  
                            dp[c^1][i+1][i+1-n]+=dp[c][i][j]*y;  
                        else  
                            dp[c^1][i+1][j]+=dp[c][i][j]*y;  
                        dp[c^1][i-1][j]+=dp[c][i][j]*x;  
                        dp[c^1][i][j]+=dp[c][i][j]*z;  
                    }  
                }  
            }  
            c=c^1;  
        }  
        for(i=0,ans=0;i<=n+n;i++)  
            for(j=1;j<=n;j++)  
            {  
                if(dp[c][i][j]>0)  
                    ans+=dp[c][i][j]*j;  
            }  
        printf("%d %.4lf\n",no,ans);  
    }  
    return 0;  
}  

#include
#include
double dp[2][205][105];
int main()
{
	int i,j,k,n,t,no,c;
	double x,y,z,ans;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%lf%lf",&no,&n,&x,&y);
		c=0;
		z=1.0-x-y;
		memset(dp[0],0,sizeof(dp[0]));
		dp[0][n][0]=1;
		for(k=0;k0)
					{
						if(i+1>j+n)
							dp[c^1][i+1][i+1-n]+=dp[c][i][j]*y;
						else
							dp[c^1][i+1][j]+=dp[c][i][j]*y;
						dp[c^1][i-1][j]+=dp[c][i][j]*x;
						dp[c^1][i][j]+=dp[c][i][j]*z;
					}
				}
			}
			c=c^1;
		}
		for(i=0,ans=0;i<=n+n;i++)
			for(j=1;j<=n;j++)
			{
				if(dp[c][i][j]>0)
					ans+=dp[c][i][j]*j;
			}
		printf("%d %.4lf\n",no,ans);
	}
	return 0;
}