uva 11722 - Joining with Friend(概率)

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题目连接：uva 11722 - Joining with Friend

题目大意：你和朋友乘火车，并且都会路过A市，给定两人可能到达A市的时段，火车会停w，问说两人可以见面的概率。

解题思路：y = x + w 和y = x - w在给定时间内围成的面积除以时间的总面积，就是求面积的时候要分情况处理。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int t1, t2, s1, s2, w;double sum;double solve (int k) {    double ret = 0;    if (s2 > t2 + k && s1 > t1 + k) {        double r = max(t2 + k - s1, 0);        ret = sum - r * r / 2;     } else if (s2 > t2 + k) {        double r = 2 * s2 - t1 - t2 - 2 * k;        ret = r * (t2 - t1) / 2;    } else if (s1 > t1 + k) {        double r = s2 + s1 - 2 * k - 2 * t1;        ret = r * (s2 - s1) / 2;    } else {        double r = max(s2 - k - t1, 0);        ret = r * r / 2;    }    if (k < 0)        ret = sum - ret;    return ret;}int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        scanf("%d%d%d%d%d", &t1, &t2, &s1, &s2, &w);        sum = 1.0 * (t2 - t1) * (s2 - s1);        double ans = solve(w) + solve(-w);        printf("Case #%d: %.8lf\n", kcas, 1 - ans / sum);    }    return 0;}

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