CF 304A(Pythagorean Theorem II-n内勾股数)

最新推荐文章于 2020-07-11 10:57:20 发布

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本文深入探讨了数学中的Pythagorean定理，并解释了如何将其应用于编程解决问题，具体通过实例展示了如何计算满足特定条件的直角三角形数量。

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In mathematics, the Pythagorean theorem — is a relation in Euclidean geometry among the three sides of a right-angled triangle. In terms of areas, it states:

In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the Pythagorean equation:

a² + b² = c²

where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides.

Given n, your task is to count how many right-angled triangles with side-lengths a, b and c that satisfied an inequality 1 ≤ a ≤ b ≤ c ≤ n.

Input

The only line contains one integer n (1 ≤ n ≤ 10⁴) as we mentioned above.

Output

Print a single integer — the answer to the problem.

Sample test(s)

input
5

output
1

input
74

output
35

显然n^2暴力枚举

注意要加优化-if (i*i+j*j>n*n) break. 不然我的n^2过不了

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cstring>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Forp(x) for(int p=pre[x];p;p=next[p])
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
    int n;
    cin>>n;
    int ans=0;
    For(i,n)
        for(int j=i+1;j<=n;j++)
        {
            int c=i*i+j*j;
            if (c>n*n) break;
            double x=sqrt(c);
            if (abs((x-(int)x))<1e-8) ans++;
        }
    cout<<ans<<endl;

	return 0;
}