数学概念——A 几何概型

You are going from Dhaka to Chittagong by train and you came to know one of your old friends is going from city Chittagong to Sylhet. You also know that both the trains will have a stoppage at junction Akhaura at almost same time. You wanted to see your friend there. But the system of the country is not that good. The times of reaching to Akhaura for both trains are not fixed. In fact your train can reach in any time within the interval [t1, t2] with equal probability. The other one will reach in any time within the interval [s1, s2] with equal probability. Each of the trains will stop for w minutes after reaching the junction.  You can only see your friend, if in some time both of the trains is present in the station. Find the probability that you can see your friend.
Input
The first line of input will denote the number of cases T (T < 500). Each of the following T line will contain 5 integers t1, t2, s1, s2, w (360 ≤ t1 < t2 < 1080, 360 ≤ s1 < s2 < 1080 and 1 ≤ w ≤ 90). All inputs t1, t2, s1, s2 and w are given in minutes and t1, t2, s1, s2 are minutes since midnight00:00.
Output
For each test case print one line of output in the format “Case #k: p” Here k is the case number and p is the probability of seeing your friend. Up to 1e-6 error in your output will be acceptable.
Sample Input	Output for Sample Input
2 1000 1040 1000 1040 20 720 750 730 760 16	Case #1: 0.75000000 Case #2: 0.67111111

解题思路：

题意：你和朋友都要乘坐火车，为了在A城市见面，你会在时间区间[t1, t2]中的任意时刻以相同的概率密度到达，你

的朋友会在时间区间[s1, s2]内的任意时刻以相同的概率密度到达，你们的火车都会在车站停留W秒，

只有在同一时刻火车都在城市A的时候，才会相见，问你这件事情的概率

思路：将密度构造成矩形，然后求解y = x +（-） b围成的图形占矩阵面积的大小，分情况讨论

程序代码：

#include <cstdio>
using namespace std;
double t1,t2,s1,s2;
double cal(double w)
{
    double x1=s1-w,x2=s2-w,y1=t1+w,y2=t2+w;
    if(y1>=s2) return ;
    if(y2<=s1) return(t2-t1)*(s2-s1);
    bool l=(y1>=s1&&y1<=s2);
    bool r=(y2>=s1&&y2<=s2);
    bool u=(x2>=t1&&x2<=t2);
    bool d=(x1>=t1&&x1<=t2);
    if(l&&u) return (x2-t1)*(s2-y1)*0.5;
    if(l&&r) return (s2-y1+s2-y2)*(t2-t1)*0.5;
    if(d&&u) return (x2-t1+x1-t1)*(s2-s1)*0.5;
    if(d&&r) return (t2-t1)*(s2-s1)-(t2-x1)*(y2-s1)*0.5;
}
int main()
{
    int Case=,t;
    double w;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf%lf",&t1,&t2,&s1,&s2,&w);
        printf("Case #%d: %.7lf\n",++Case,(cal(-w)-cal(w))/(t2-t1)/(s2-s1));
    }
    return ;
}