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2014 BNU邀请赛F题(枚举)

Football on Table

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?/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;
}