UVA 11722 Joining with Friend [连续概率]

　　A在[S1,S2]到达,B在[T1,T2]到达,到达后都会等W分钟，问相遇的概率。

　　概率论课本上的题目了,列出三个方程s1<=S<=s2,t1<=T<=t2,|T-S|<=w。然后可以转化为几何图形求面积，需要讨论求解。

 1 #include <string.h>
 2 #include <stdio.h>
 3 int cas, t1, t2, s1, s2, w;
 4 double area(int w){
 5     int lc = t1+w, rc = t2+w, uc = s2-w, dc = s1-w;
 6     if (lc >= s2) return 0;
 7     if (rc <= s1) return (t2-t1)*(s2-s1);
 8     bool bl = (lc>=s1 && lc<=s2);
 9     bool br = (rc>=s1 && rc<=s2);
10     bool bu = (uc>=t1 && uc<=t2);
11     bool bd = (dc>=t1 && dc<=t2);
12     if (bl&&bu) return (lc-s2)*(lc-s2)*0.5;
13     if (bl&&br) return (t2-t1)*(s2-lc+s2-rc)*0.5;
14     if (bd&&bu) return (s2-s1)*(uc-t1+dc-t1)*0.5;
15     if (bd&&br) return (t2-t1)*(s2-s1)-(rc-s1)*(rc-s1)*0.5;
16     return 0;
17 }
18 int main(){
19     scanf("%d", &cas);
20     for (int ca = 1; ca <= cas; ca++) {
21         scanf("%d%d%d%d%d", &t1, &t2, &s1, &s2, &w);
22         printf("Case #%d: %.7f\n", ca, (area(-w)-area(w))/(t2-t1)/(s2-s1));
23     }
24     return 0;
25 }