HDOJ题目2105The Center of Gravity

本文介绍了一个简单的算法,用于计算由三个点定义的三角形的重心坐标。通过输入三个顶点的坐标,可以精确到小数点后一位地计算并输出重心坐标。

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The Center of Gravity

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4112    Accepted Submission(s): 2371


Problem Description
Everyone know the story that how Newton discovered the Universal Gravitation. One day, Newton walked
leisurely, suddenly, an apple hit his head. Then Newton discovered the Universal Gravitation.From then
on,people have sovled many problems by the the theory of the Universal Gravitation. What's more, wo also
have known every object has its Center of Gravity.
Now,you have been given the coordinates of three points of a triangle. Can you calculate the center
of gravity of the triangle?
 

Input
The first line is an integer n,which is the number of test cases.
Then n lines follow. Each line has 6 numbers x1,y1,x2,y2,x3,y3,which are the coordinates of three points.
The input is terminated by n = 0.
 

Output
For each case, print the coordinate, accurate up to 1 decimal places.
 

Sample Input
  
2 1.0 2.0 3.0 4.0 5.0 2.0 1.0 1.0 4.0 1.0 1.0 5.0 0
 

Sample Output
  
3.0 2.7 2.0 2.3
 

Source
 ac代码
#include<stdio.h>
#include<math.h>
int main()
{
	int t;
	while(scanf("%d",&t)!=EOF,t)
	{
	while(t--)
	{
		double x1,x2,x3,y1,y2,y3;
		scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
		printf("%.1lf %.1lf\n",(x1+x2+x3)/3.0,(y1+y2+y3)/3.0);
	}
	}
}


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