Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem D: Laser Pointer

Jake and his friends have been walking around a house of mirrors at a carnival for quite some time now. While Jake's friends are still interested in seeing a few rooms, Jake is quite bored. He decides to wait by the exit while the rest of them search the remainder of the house. Jake also has a laser pointer which he plays with to help pass the time.

The exit is located at the end of a hallway of length L and width W where 0 < L,W ≤ 10,000. The walls along the length of the hallway are lined with mirrors. Jake is at the start of the hallway and is leaning against the right wall. He shines his laser pointer toward the exit making an angle (in degrees) of 0 ≤ θ < 90 with the right wall.

The laser beam travels some distance A, perhaps reflecting off one or more walls, until it illuminates a spot at the end of the hall on the exit door. In his boredom, Jake wonders how much further this beam travelled than if he had shone the laser pointer directly at the spot on the door. That is, if the Euclidean distance from the laser pointer to the illuminated point on the wall is B, Jake is trying to determine the ratio A/B.

The first line of input indicates how many test cases follow. Each test case consists of three integers L, W, and θ on one line.

The output for each test case is a single line containing the value A/B rounded to three decimal places.

Sample input

2
10 5 45
12 12 75

Output for sample input

1.414
3.732

---

Zachary Friggstad

將圖形補滿擴充，反彈的題目都是這麼去解的。
但要考慮反彈上下面得奇偶數的數量，才能去決定離地面的高度到底是 x 還是 w-x。

#include <stdio.h>
#include <math.h>
using namespace std;
int main() {
    int testcase;
    double pi = acos(-1);
    double L, W, theta;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%lf %lf %lf", &L, &W, &theta);
        theta = theta/180*pi;
        double A, B;
        A = L/cos(theta);
        B = A*sin(theta);
        int cross = (int)(B/W);
        B = fmod(B, W);
        if(cross&1)
            B = W-B;
        B = sqrt(B*B+L*L);
        printf("%.3lf\n", A/B);
    }
    return 0;
}