给定周长P,任务是找到周长等于p的直角三角形的数量。
例子:
Input: P = 12
Output: number of right triangles = 1
The only right angle possible is with sides
hypotenuse = 5, perpendicular = 4 and base = 3.
Input: p = 840
Output: number of right triangles = 8
因此,目的是找到满足方程a + b + c = p和a 2 + b 2 = c 2的解的数量。
天真的方法是对a(1到p / 2)和b(a + 1到p / 3)运行两个循环,然后使c = pab,如果 。这将需要时间。
很少的代数运算可以找到一种有效的方法:
由于a + c> b或,p – b> b或b
下面是上述方法的实现。C++
// C++ program to find the number of
// right triangles with given perimeter
#include
using namespace std;
// Function to return the count
int countTriangles(int p)
{
// making a list to store (a, b) pairs
vector> store;
// no triangle if p is odd
if (p % 2 != 0)
return 0;
else
{
int count = 1;
for(int b = 1; b < p / 2; b++)
{
float a = (float)p / 2.0f * ((float)((float)p -
2.0 * (float)b) /
((float)p - (float)b));
int inta = (int)(a);
if (a == inta)
{
// make (a, b) pair in sorted order
pair ab;
if(intaPython3
# python program to find the number of
# right triangles with given perimeter
# Function to return the count
def countTriangles(p):
# making a list to store (a, b) pairs
store =[]
# no triangle if p is odd
if p % 2 != 0 : return 0
else :
count = 0
for b in range(1, p // 2):
a = p / 2 * ((p - 2 * b) / (p - b))
inta = int(a)
if (a == inta ):
# make (a, b) pair in sorted order
ab = tuple(sorted((inta, b)))
# check to avoid duplicates
if ab not in store :
count += 1
# store the new pair
store.append(ab)
return count
# Driver Code
p = 840
print("number of right triangles = "+str(countTriangles(p)))
输出:
number of right triangles = 8
时间复杂度: O(P)