HDOJ 3232 simple probability of Crossing Rivers,

HDOJ 3232 simple probability of Crossing Rivers,

Simple expectation:

The expected time for the ship to the shore is that the time for the ship to cross the river is L/v.

Therefore, the expected time for crossing each river is 2 * L/v.

Crossing Rivers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 699 Accepted Submission (s): 361


Problem DescriptionYou 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 ExpectedTime to reach B from? Your walking speed is always 1.

To be more precise, for a river of length L, the distance of the boat (which cocould be regarded as a mathematical point) to the left bank at time 0 is Uniformly chosenFrom interval [0, L], And the boat is equally like to be moving left or right, if it's not precisely at the river bank.
InputThere will be at most 10 test cases. Each case begins with two integers NAnd 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 NLines describes a river with 3 integers: P, LAnd V(0 <= P< D, 0 < L<= D, 1 <= V<= 100 ). PIs the distance from A to the left bank of this river, LIs the length of this river, VIs 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 N = D = 0, Which shoshould not be processed.
OutputFor 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 10 1 20 10 0

 
Sample Output

Case 1: 1.000Case 2: 1.000

 
Source2009 Asia Wuhan Regional Contest Hosted by Wuhan University

/*************************************** * ******** Author: CKbossCreated Time:, Thursday, January 29, 2015 File Name: UVA12230.cpp *************************************** * ********/# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <string> # include <cmath> # include <cstdlib> # include <vector> # include <queue> # include <set> # include <map> using namespace std; int n, d; struct Riv Er {int p, l, v;} ri [20]; int main () {// freopen ("in.txt", "r", stdin ); // freopen ("out.txt", "w", stdout); int cas = 1; while (scanf ("% d", & n, & d )! = EOF) {if (n = 0 & d = 0) break; double ans1 = 0.0; double sum = 0.0; for (int I = 0; I <n; I ++) {int a, B, c; scanf ("% d", & a, & B, & c ); ri [I] = (River) {a, B, c}; ans1 + = 2. * B/c; sum + = B;} ans1 + = (d-sum); printf ("Case % d: %. 3lf \ n ", cas ++, ans1); putchar (10);} return 0 ;}