hdu 3232 Crossing Rivers

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.

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【分析】
求走过d路程的时间的期望值
对于每条河流，过河的时间在l/v和3l/v之间均匀分布，所以期望时间是2l/v。

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【代码】

//hdu 3232 Crossing Rivers #include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define M(a) memset(a,0,sizeof a)#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;int n,t;double d;int main(){    while(scanf("%d%lf",&n,&d) && (n||d))    {        double ans=0,sum=0,v,x,l;        t++;        while(n--)        {            scanf("%lf%lf%lf",&x,&l,&v);            ans-=l;            ans+=2*l/v;        }        ans+=d;        printf("Case %d: %.3lf\n\n",t,ans);    }    return 0;}