"Bored?

Let‘s play table football!"

The table football is played on a rectangular table, usually contains m rows of players which are plastic, metal, wooden, or sometimes carbon-fibre figures mounted on vertical metal bars. After playing table football for hours, we decide to take a rest. And the state of the table remains random, that means each bar is placed at any legal position with equal possibilities (players can’t be outside the table and a bar is fixed at a row).

 

 

Now I‘m wondering if the goal-keeper shoot a ball, what’s the possibility of this shoot turning to a goal?

(If the ball did not touch any player, then I made a goal).

Let‘s assume there is a_i players on the i^th row (counted from left to right). And we know the width of each player and the distance between two players. (To simplify the problem, we ignore the thickness of the players, in other words, we consider the players as vertical segments. Then we treat the football as a point, moving along a straight line and will not touch the boundary of the table).

 

Input

Output

Sample Input

2
8.0 10.0
0.0 5.0 2.0 -0.1
1
3.0 2
2.0 2.0
1.0
8.0 10.0
0.0 5.0 2.0 0.0
2
3.0 2
2.0 2.0
1.0
4.0 3
2.0 1.0 2.0
1.0 1.0

Sample Output

Case #1: 0.23000
Case #2: 0.13333

Hint

The black solid lines denote the table.

The dashed line denotes the bar.

The gray lines denote the players.

The dot-dashed line denote the trajectory of the ball.

 

Source

题意：：一个人在玩桌面足球，有m行球员。每行球员有ai个，给出每一个球员的宽度和相邻球员之间的距离，球从最左边射出，给出球的起点坐标跟方向向量，问可以到达最右边的概率。

思路：看懂题意就好做点了，枚举每行能够碰到球的距离。然后概率求反

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 105;

double w[maxn], tail[maxn], pos[maxn];

int main() {
	int t, m, cas = 1;
	double W, L, X, Y, dx, dy, dis;
	scanf("%d", &t);
	while (t--) {
		double ans = 1.0;
		scanf("%lf%lf", &L, &W);
		scanf("%lf%lf%lf%lf", &X, &Y, &dx, &dy);
		scanf("%d", &m);

		while (m--) {
			double sum = 0.0;
			pos[0] = 0.0;
			double x;
			int n;
			scanf("%lf%d", &x, &n);
			double y = Y + dy * (x - X) / dx;
			for (int i = 0; i < n; i++) {
				scanf("%lf", &w[i]);
				sum += w[i];
			}
			tail[0] = w[0];
			for (int i = 1; i < n; i++) {
				scanf("%lf", &dis);
				pos[i] = pos[i-1] + w[i-1] + dis;
				tail[i] = pos[i] + w[i];
				sum += dis;
			}
			double cnt = 0.0, len = 0.0;
			double mv = W - sum;
			for (int i = 0; i < n; i++) {
				cnt = 0.0;
				if ((pos[i] <= y) && (tail[i] + mv) >= y) {
					if (tail[i] >= y) 
						cnt = (pos[i] + mv <= y) ? mv : (y - pos[i]);
					else cnt = (pos[i] + mv >= y) ?
 w[i] : (tail[i] - y + mv);
				}
				len += cnt;
			}
			if (mv == 0.0) {
				ans = 0;
				break;
			}
			else ans = ans * (mv - len) / mv;
		}

		printf("Case #%d: %.5lf\n", cas++, ans);
	}
	return 0;
}