题意:一些杆上有人,人有一个宽度,然后现在有一个球射过去,要求出球不会碰到任何人的概率
思路:计算出每根杆的概率,之后累乘,计算杆的概率的时候,可以先把每块人的区间长度再移动过程中会覆盖多少长度累加出来,然后(1?总和/可移动距离)就是不会碰到的概率
代码:
#include <stdio.h>
#include <string.h>
#include <math.h>
const double eps = 1e-8;
int t, n;
double l, w, x, y, dx, dy;
struct gan {
double len;
int num;
double r[105];
double d[105];
double sumd;
void init() {
sumd = 0;
memset(r, 0, sizeof(r));
memset(d, 0, sizeof(d));
}
} g;
double cal(gan g) {
double yy = g.len * dy / dx + y;
double yi = w - g.sumd;
double down = 0, up = g.r[0];
double ans = 0;
for (int i = 0; i < g.num; i++) {
if (up + yi > yy && down < yy) {
if (up > yy) {
if (down + yi > yy) ans += yy - down;
else ans += yi;
}
else {
if (down + yi > yy) ans += g.r[i];
else ans += yi - (yy - up);
}
}
down = up + g.d[i];
up = down + g.r[i + 1];
}
return 1.0 - ans / yi;
}
int main() {
int cas = 0;
scanf("%d", &t);
while (t--) {
double ans = 1;
scanf("%lf%lf", &l, &w);
scanf("%lf%lf%lf%lf", &x, &y, &dx, &dy);
scanf("%d", &n);
for (int i = 0; i < n; i++) {
g.init();
scanf("%lf%d", &g.len, &g.num);
for (int j = 0; j < g.num; j++) {
scanf("%lf", &g.r[j]);
g.sumd += g.r[j];
}
for (int j = 0; j < g.num - 1; j++) {
scanf("%lf", &g.d[j]);
g.sumd += g.d[j];
}
g.d[g.num - 1] = 0;
ans *= cal(g);
}
printf("Case #%d: %.5lf\n", ++cas, ans);
}
return 0;
}
原文地址:http://blog.csdn.net/wangxiaobupt/article/details/26734817