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 C	Campus Roads	Time Limit : 2 seconds
	NGU has a beautiful campus. The authority of NGU has decided to take some more steps about beautification of this campus. One of these steps is tree plantation in some roads. To complete this project they will select some roads, and for any road they will also select a number that how many tree will be planted on that road. To make this campus more beautiful they decided in a single road, the gap between any two adjacent tree will be fixed, and this gap will be as large as possible. So If you walk through the road in a constant speed, you will reach a tree after constant intervals. A road can be described by K points P1, P2, ...... , PK. Pi and Pi+1 are connected by a straight line. So, it is clear that a road consists of some contiguous straight lines.   In this problem you have to find all the points, where to plant a tree. For the sake of simplicity, we will consider that roads will not have any width.
	Input
	The first line of input is an integer N (N<100) that indicates the number of roads under this beautification project. From next line N roads is described one by one. Description of each road starts with a line containing two integers K (2≤K≤100) and T (2≤T≤100).  Here K represents number of points that represents the road, and T represents the number of trees to be planted. This line is followed by K lines having 2 integers x, y(-1000.00<=x<=1000.00 and -1000.00<=y<=1000.00) each, ith line of these K lines is the co-ordinate of  Pi.
	Output
	You have to give your output for every roads one by one. Output of each road starts with a line "Road #n:" (without quotes), here n is the road number. Next T lines will give the co-ordinates of  trees to be planted. x, y co-ordinate will be separated by a single space and having two decimal points. You have to represent the points sequentially in the order in which order you will find those trees when you walk through the road P1 to PK. Print a blank line after describing each road.
	Sample Input	Sample Output
	1 5 6 10.00 10.00 20.00 20.00 30.00 10.00 10.00 0.00 9.00 9.00	Road #1: 10.00 10.00 18.44 18.44 26.89 13.11 23.26 6.63 12.58 1.29 9.00 9.00
	Hint: You have to maximize the distance between two adjacent trees in a road. So it is sure that first and last tree will be on the first and last point, i.e. in P1 and Pk. Problem Setter: Md. Arifuzzaman Arif Special Thanks: Shamim Hafiz Next Generation Contest 5

根據走的路線，每隔 ? 公尺，放置一棵樹，特別注意，可能一個線段上一次放置好幾棵樹。

#include <stdio.h>
#include <math.h>
#define eps 1e-8
int main() {
    int t, n, m, cases = 0, i, j;
    double x[105], y[105];
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        for(i = 0; i < n; i++)
            scanf("%lf %lf", x+i, y+i);
        printf("Road #%d:\n", ++cases);
        double path = 0, sum = 0, pre = 0;
        for(i = 1; i < n; i++)
            path += sqrt(pow(x[i]-x[i-1], 2)+pow(y[i]-y[i-1],2));
        path /= m-1;
        printf("%.2lf %.2lf\n", x[0], y[0]);
        for(i = 1; i < n; i++) {
            double len = sqrt(pow(x[i]-x[i-1], 2)+pow(y[i]-y[i-1],2));
            double sx = x[i-1], sy = y[i-1], ex = x[i], ey = y[i];
            pre = sum;
            if(sum+len >= path-eps) {
                double d = path - sum;
                sum = 0;
                sx = sx + d/len*(ex-sx);
                sy = sy + d/len*(ey-sy);
                printf("%.2lf %.2lf\n", sx, sy);
                len -= d;
                d = path;
                while(len >= path-eps) {
                    sx = sx + d/len*(ex-sx);
                    sy = sy + d/len*(ey-sy);
                    printf("%.2lf %.2lf\n", sx, sy);
                    len -= path;
                }
            }
            sum += len;
        }
        puts("");
    }
    return 0;
}