标签：sicily   几何   

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Problem Cylinders (e.g. oil drums) (of radius 1 foot) are stacked in a rectangular bin. Each cylinder on an upper row rests on two cylinders in the row below. The cylinders in the bottom row rest on the floor and do not roll from their original positions. Each row has one less cylinder than the row below.

This problem is to write a program to compute the location of the center of the top cylinder from the centers of the cylinders on the bottom row. Computations of intermediate values should use double precision.

Input

The input begins with a line containing the count of problem instances, nProb , as a decimal integer, (1<=nProb<=1000) . This is followed by nProb input lines. An input line consists of the number, n , of cylinders on the bottom row followed by n floating point values giving the x coordinates of the centers of the cylinders (the y coordinates are all 1.0 since the cylinders are resting on the floor (y = 0.0 )). The value of n will be between 1 and 10 (inclusive). The distance between adjacent centers will be at least 2.0 (so the cylinders do not overlap) and at most 3.4 (so cylinders at level k cannot touch cylinders at level k - 2 ).

Output

The output for each data set is a line containing the problem number (1...nProb) , a colon, a space, the x coordinate of the topmost cylinder to 4 decimal places, a space and the y coordinate of the topmost cylinder to 4 decimal places (both x and y ROUNDED to the nearest digit).

Note: To help you check your work, the x -coordinate of the center of the top cylinder should be the average of the x -coordinates of the leftmost and rightmost bottom cylinders. 

Sample Input

5 
4 1.0 4.4 7.8 11.2 
1 1.0 
6 1.0 3.0 5.0 7.0 9.0 11.0 
10 1.0 3.0 5.0 7.0 9.0 11.0 13.0 15.0 17.0 20.4 
5 1.0 4.4 7.8 11.2 14.6

Sample Output

1: 6.1000 4.1607 
2: 1.0000 1.0000 
3: 6.0000 9.6603 
4: 10.7000 15.9100 
5: 7.8000 5.2143

假设最底层有N个球，那么每↑一层，则有N-1个球，最顶层为1个球。思路很简单，先把底层的球按x坐标从小到大排一次序，然后每相邻的两个球求出叠在它们上面的那个球的坐标(这里用到点三角函数的知识，自己在纸上演算一下并不难得出坐标关系)。

已经两个圆，求一个跟它们相切的圆，我们可以构造一个等腰三角形，腰为2。

这样时间复杂度是O(n^2)

#include <bits/stdc++.h>
using namespace std;
double x[11];//一共 
double y[11];

void solve(double x1,double y1,double x2,double y2,double &x3,double &y3){
	double len=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
	double mid=sqrt(4-len*len/4);
	double mid_x=(x1+x2)/2;
	double mid_y=(y1+y2)/2;
	double cos_=(x2-x1)/len;
	double sin_=(y1-y2)/len;
	x3=sin_*mid+mid_x;
	y3=cos_*mid+mid_y;
}
int main(){
	int t;
	scanf("%d",&t);
	int tmp=t;
	while(t--){
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			scanf("%lf",&x[i]);
			y[i]=1; 
		}
		sort(x+1,x+n+1);
		for(int i=n;i>=2;i--){
			for(int j=1;j<i;j++)
			solve(x[j],y[j],x[j+1],y[j+1],x[j],y[j]);
		}	
		printf("%d: %.4lf %.4lf\n",tmp-t,x[1],y[1]);
	}
} 

sicily 1012/1206 Stacking Cylinders

标签：sicily   几何   

原文地址：http://blog.csdn.net/yuhao199555/article/details/46328049