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: Cutting tabletops

Bever Lumber hires beavers to cut wood. The company has recently received a shippment of tabletops. Each tabletop is a convex polygon. However, in this hard economic times of cutting costs the company has ordered the tabletops from a not very respectable but cheap supplier. Some of the tabletops have the right shape but they are slightly too big. The beavers have to chomp of a strip of wood of a fixed width from each edge of the tabletop such that they get a tabletop of a similar shape but smaller. Your task is to find the area of the tabletop after beavers are done.

Input consists of a number of cases each presented on a separate line. Each line consists of a sequence of numbers. The first number is d the width of the strip of wood to be cut off of each edge of the tabletop in centimeters. The next number n is an integer giving the number of vertices of the polygon. The next n pairs of numbers present x_i and y_i coordinates of polygon vertices for 1 <= i <= n given in clockwise order. A line containing only two zeroes terminate the input.

d is much smaller than any of the sides of the polygon. The beavers cut the edges one after another and after each cut the number of vertices of the tabletop is the same.

For each line of input produce one line of output containing one number to three decimal digits in the fraction giving the area of the tabletop after cutting.

Sample input

2 4 0 0 0 5 5 5 5 0
1 3 0 0 0 5 5 0
1 3 0 0 3 5.1961524 6 0
3 4 0 -10 -10 0 0 10 10 0
0 0

Output for sample input

1.000
1.257
2.785
66.294

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Problem Setter: Piotr Rudnicki

給一個凸多邊形, 且向內部切割一個距離 d。

先計算該多邊形的面積，接著計算夾角的三角形夾角 theta，
theta 對應的是直角三角形的邊 d, 而這個三角形有兩個, 計算其面積。

而矩形的周長會是原本多邊形的周長減去三角形另一個邊的長度。


#include <stdio.h>
#include <math.h>
struct Pt {
    double x, y;
};
int main() {
    const double pi = acos(-1);
    double d;
    int n, i, j;
    Pt D[1005];
    while(scanf("%lf %d", &d, &n) == 2 && n) {
        for(i = 0; i < n; i++)
            scanf("%lf %lf", &D[i].x, &D[i].y);
        D[n] = D[0];
        double area = 0;
        for(i = 0; i < n; i++)
            area += D[i].x*D[i+1].y-D[i].y*D[i+1].x;
        area = fabs(area)/2;
        double perimeter = 0;
        for(i = 0; i < n; i++)
            perimeter += hypot(D[i].x-D[i+1].x, D[i].y-D[i+1].y);
        D[n+1] = D[1];
        for(i = 1; i <= n; i++) {
            double a, b, c;
            a = hypot(D[i].x-D[i-1].x, D[i].y-D[i-1].y);
            b = hypot(D[i].x-D[i+1].x, D[i].y-D[i+1].y);
            c = hypot(D[i+1].x-D[i-1].x, D[i+1].y-D[i-1].y);
            double theta = acos((a*a+b*b-c*c)/(2*a*b));
            theta /= 2;
            double w = d/tan(theta);
            perimeter -= w+w;
            area -= d*w;
        }
        area -= perimeter*d;
        printf("%.3lf\n", area);
    }
    return 0;
}