HDU 3232Crossing Rivers(数学期望)

Problem Description
You live in a village but work in another village. You decided to follow the straight path between your house (A) and the working place (B), but there are several rivers you need to cross. Assume B is to the right of A, and all the rivers lie between them.
Fortunately, there is one "automatic" boat moving smoothly in each river. When you arrive the left bank of a river, just wait for the boat, then go with it. You're so slim that carrying you does not change the speed of any boat.
Days and days after, you came up with the following question: assume each boat is independently placed at random at time 0, what is the expected time to reach B from A? Your walking speed is always 1.
To be more precise, for a river of length L, the distance of the boat (which could be regarded as a mathematical point) to the left bank at time 0 is uniformly chosen from interval [0, L], and the boat is equally like to be moving left or right, if it’s not precisely at the river bank.
Input
There will be at most 10 test cases. Each case begins with two integers n and D, where n (0 <= n <= 10) is the number of rivers between A and B, D (1 <= D <= 1000) is the distance from A to B. Each of the following n lines describes a river with 3 integers: p, L and v (0 <= p < D, 0 < L <= D, 1 <= v <= 100). p is the distance from A to the left bank of this river, L is the length of this river, v is the speed of the boat on this river. It is guaranteed that rivers lie between A and B, and they don’t overlap. The last test case is followed by n=D=0, which should not be processed.
Output
For each test case, print the case number and the expected time, rounded to 3 digits after the decimal point.
Print a blank line after the output of each test case.
Sample Input
1 1
0 1 2
0 1
0 0
Sample Output
Case 1: 1.000

Case 2: 1.000

AC代码：

这题是这个样子，首先你到达河边的时候， 由于船的位置和方向是任意的，你离船的距离平均为L/2，假如船这时候面对你，那么船开过来只要L/2的距离，如果是背对你，那么开过来的距离就是3*L/2，平均过来的距离就是L；所以总的过河距离就是2L；

# include <cstdio>using namespace std;//由于船的方向不确定位置不确定，要行使的距离大约为河长的两倍 double cal(int l, int v){return double(l)*2/v;}int main(){int n, d, i, j, k, cnt=0, p, l, v;double ans, sum;while(scanf("%d%d", &n, &d)){if(n==0&&d==0){break;}sum=0;ans=0.0;for(i=1; i<=n; i++){scanf("%d%d%d", &p, &l, &v);ans=ans+cal(l, v);sum=sum+l;}ans=ans+d-sum;if(!(n==0&&d==0))printf("Case %d: %.3lf\n\n", ++cnt, ans);}return 0;}