UVA 12230 – crossing rivers (probability) "suggested collection"

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Topic link ：uva 12230 – Crossing Rivers

The main idea of the topic ： A man goes to work in the company every day . Every time I pass N Rivers . The distance between home and company is D. The default speed on land is 1, give N Information about the river . Include starting coordinates p, Width L, And the speed of the ship .

The boat will go back and forth on both sides of the river . People reach the river bank is . The position of the ship is random （ Including direction ）. Ask about the expected time for people to reach the company .

Their thinking ： Time on land is fixed , Just calculate the time of each river separately . Because when people get to the shore , The position of the ship is random , So the waiting time [0,2L/v], The period is equal probability , So the time to cross a river is extremely (0+2∗Lv)2+Lv=2∗Lv

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int main () {
    int cas = 1;
    int N;
    double D, p, l, v;

    while (scanf("%d%lf", &N, &D) == 2 && (N || D)) {

        for (int i = 0; i < N; i++) {
            scanf("%lf%lf%lf", &p, &l, &v);
            D = D - l + 2 * l / v;
        }

        printf("Case %d: %.3lf\n\n", cas++, D);
     }
    return 0;
}

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