题目连接：bnu 34986 Football on Table

题目大意:给出桌子的大小L，W，然后是球的起始位置sx，sy，以及移动的向量dx，dy，然后给出n，表示有n个杆，对于每个杆，先给出位置x，以及杆上有多少个小人c，给出小人的宽度，再给出c个小人间的距离。现在问说球有多少个概率可以串过所有人。

解题思路；对于每个杆求无阻挡的概率，注意概率 = 空隙 / 可移动的范围大小，而不是W。其他就水水的。

#include 
   
     #include 
    
      #include 
     
       #include 
      
        using namespace std; const double eps = 1e-8; const int N = 205; double L, W; double sx, sy, dx, dy; double d[N]; double solve () { int n, c; double w, ans = 1; scanf("%d", &n); while (n--) { scanf("%lf%d", &w, &c); double r = sy + w * dy / dx; double s = 0; c = 2 * c - 1; for (int i = 0; i < c; i += 2) scanf("%lf", &d[i]); for (int i = 1; i < c; i += 2) scanf("%lf", &d[i]); for (int i = 0; i < c; i++) s += d[i]; double l = W - r; double rec = 0; if (l > s) { rec += (l - s); l = 0; } else { l = s - l; } if (r > s) { rec += (r - s); r = s; } s = 0; for (int i = 0; i < c; i++) { double tmp = s + d[i]; if (i&1) { double add = min(r, tmp) - max(s, l); rec += max(add, (double)0); } s = tmp; } ans *= rec / (W-s); } return ans; } int main () { int cas; scanf("%d", &cas); for (int i = 1; i <= cas; i++) { scanf("%lf%lf", &L, &W); scanf("%lf%lf%lf%lf", &sx, &sy, &dx, &dy); printf("Case #%d: %.5lf\n", i, solve()); } return 0; }