使用边长的循环四边形的对角线长度。
给定整数 A 、 B 、 C 和 D ，表示一个循环四边形的边长，任务是求一个循环四边形的对角线长度。
示例:
输入: A = 10，B = 15，C = 20，D = 25 T3】输出: 22.06 26.07
输入: A = 10，B = 30，C =50，D = 20 T3】输出: 37.93 29.0
方法:对角线的长度可以用以下公式计算:
【tex】对角(q)= \ sqrt {(AC+BD)(ab+CD } } { ad+BC } }[/tex]
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the length of
// diagonals of a cyclic quadrilateral
vector<float> Diagonals(int a, int b,
int c, int d)
{
vector<float> ans;
ans.push_back(sqrt(((a * c) + (b * d)) *
((a * d) + (b * c)) /
((a * b) + (c * d))));
ans.push_back(sqrt(((a * c) + (b * d)) *
((a * b) + (c * d)) /
((a * d) + (b * c))));
return ans;
}
// Driver Code
int main()
{
int A = 10;
int B = 15;
int C = 20;
int D = 25;
// Function Call
vector<float> ans = Diagonals(A, B, C, D);
// Print the final answer
printf("%.2f %.2f",
(ans[0]) + .01,
ans[1] + .01);
}
// This code is contributed by Amit Katiyar
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to implement
// the above approach
import java.util.*;
class GFG{
// Function to calculate the length of
// diagonals of a cyclic quadrilateral
static Vector<Float> Diagonals(int a, int b,
int c, int d)
{
Vector<Float> ans = new Vector<Float>();
ans.add((float) Math.sqrt(((a * c) + (b * d)) *
((a * d) + (b * c)) /
((a * b) + (c * d))));
ans.add((float) Math.sqrt(((a * c) + (b * d)) *
((a * b) + (c * d)) /
((a * d) + (b * c))));
return ans;
}
// Driver Code
public static void main(String[] args)
{
int A = 10;
int B = 15;
int C = 20;
int D = 25;
// Function Call
Vector<Float> ans = Diagonals(A, B,
C, D);
// Print the final answer
System.out.printf("%.2f %.2f",
(ans.get(0)) + .01,
ans.get(1) + .01);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program to implement
# the above approach
import math
# Function to calculate the length of
# diagonals of a cyclic quadrilateral
def Diagonals(a, b, c, d):
p = math.sqrt(((a * c)+(b * d))*((a * d)+(b * c))
/ ((a * b)+(c * d)))
q = math.sqrt(((a * c)+(b * d))*((a * b)+(c * d))
/ ((a * d)+(b * c)))
return [p, q]
# Driver Code
A = 10
B = 15
C = 20
D = 25
# Function Call
ans = Diagonals(A, B, C, D)
# Print the final answer
print(round(ans[0], 2), round(ans[1], 2))
C
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to calculate the length of
// diagonals of a cyclic quadrilateral
static List<float> Diagonals(int a, int b,
int c, int d)
{
List<float> ans = new List<float>();
ans.Add((float) Math.Sqrt(((a * c) + (b * d)) *
((a * d) + (b * c)) /
((a * b) + (c * d))));
ans.Add((float) Math.Sqrt(((a * c) + (b * d)) *
((a * b) + (c * d)) /
((a * d) + (b * c))));
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int A = 10;
int B = 15;
int C = 20;
int D = 25;
// Function Call
List<float> ans = Diagonals(A, B,
C, D);
// Print the readonly answer
Console.Write("{0:F2} {1:F2}",
(ans[0]) + .01,
ans[1] + .01);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript Program to implement
// the above approach
// Function to calculate the length of
// diagonals of a cyclic quadrilateral
function Diagonals(a, b, c, d)
{
var p = parseFloat(
Math.sqrt(((a * c) + (b * d)) *
((a * d) + (b * c)) /
((a * b) + (c * d))));
var q = parseFloat(
Math.sqrt(((a * c) + (b * d)) *
((a * b) + (c * d)) /
((a * d) + (b * c))));
return [p, q];
}
// Driver Code
var A = 10;
var B = 15;
var C = 20;
var D = 25;
// Function Call
var ans = Diagonals(A, B, C, D)
// Print the final answer
document.write(ans[0].toFixed(2) + " ",
ans[1].toFixed(2));
// This code is contributed by kirti
</script>
Output:
22.06 26.07
时间复杂度:O(1) T5辅助空间:** O(1)
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