hdu 4386Quadrilateral

最新推荐文章于 2022-08-29 19:46:08 发布

转载最新推荐文章于 2022-08-29 19:46:08 发布 · 114 阅读

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原文链接：http://www.cnblogs.com/1114250779boke/archive/2012/11/09/2763205.html

本文介绍了一道编程题，要求计算由四根给定长度棍子构成的最大面积四边形。利用圆内接四边形特性，通过计算半周长来得出最大面积公式，并提供了解题代码。

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http://acm.hdu.edu.cn/showproblem.php?pid=4386

Quadrilateral

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 937 Accepted Submission(s): 421

Problem Description

　　One day the little Jack is playing a game with four crabsticks. The game is simple, he want to make all the four crabsticks to be a quadrilateral, which has the biggest area in all the possible ways. But Jack’s math is so bad, he doesn’t know how to do it, can you help him using your excellent programming skills?

Input

　　The first line contains an integer N (1 <= N <= 10000) which indicates the number of test cases. The next N lines contain 4 integers a, b, c, d, indicating the length of the crabsticks.(1 <= a, b, c, d <= 1000)

Output

　　For each test case, please output a line “Case X: Y”. X indicating the number of test cases, and Y indicating the area of the quadrilateral Jack want to make. Accurate to 6 digits after the decimal point. If there is no such quadrilateral, print “-1” instead.

Sample Input

   2 1 1 1 1 1 2 3 4
  

Sample Output

   Case 1: 1.000000 Case 2: 4.898979
  

Author

WHU

四边形不稳定，给定四边后不能确定面积。面积最大的是圆内接四边形，设四边长为abcd，半周长为p，则最大面积=sqrt（(p-a)*(p-b)*(p-c)*(p-d))。。忘了怎么证明去了。

View Code

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include<math.h>
 4 int main()
 5 {
 6       int t;
 7       int a,b,c,d,max,count=0;
 8       double p,ans;
 9       scanf("%d",&t);
10       while(t--)
11       {
12             scanf("%d%d%d%d",&a,&b,&c,&d);
13             if(a>b) max=a;
14             else max=b;
15             if(max<b) max=b;
16             if(max<c) max=c;
17             if(max<d) max=d;
18             if(max-(a+b+c+d-max)>=0)
19             {
20                   printf("Case %d: -1\n",++count);
21                   continue;
22             }
23             p=(a+b+c+d)*1.0/2;
24             ans=sqrt((p-a)*(p-b)*(p-c)*(p-d));
25             printf("Case %d: %.6lf\n",++count,ans);
26 
27 
28       }
29 }

转载于:https://www.cnblogs.com/1114250779boke/archive/2012/11/09/2763205.html