HDU 4386 Quadrilateral（数学啊）

Posted 2020-09-19 jzssuanfa

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

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

 

Source

2012 Multi-University Training Contest 9

题意：

给出四条边的长度，求是否能形成四边形。假设能形成求最大面积。

PS：

四边形最大面积：

L = (A+B+C+D)/2；

AREA = sqrt((L-A) * (L-B)*(L-C)*(L-D));

代码例如以下：

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
    int t;
    int cas = 0;
    scanf("%d",&t);
    while(t--)
    {
        int a[4];
        for(int i = 0; i < 4; i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a,a+4);
        int sum = a[0]+a[1]+a[2];
        if(sum <= a[3])
        {
            printf("Case %d: -1\n",++cas);
            continue;
        }
        double p = (a[0]+a[1]+a[2]+a[3])/2.0;
        double area = sqrt((p-a[0])*(p-a[1])*(p-a[2])*(p-a[3]));
        printf("Case %d: %.6lf\n",++cas,area);

    }
    return 0;
}