FZU 1669 Right-angled Triangle 毕达哥拉斯三元组

 【 毕达哥拉斯三元组】

                                  X^2 + Y^2 = Z^2
    满足这个方程的的正整数三元组被称为毕达哥拉斯三元组。
    本原的毕达哥拉斯三元组，指如果一个毕达哥拉斯三元组x,y,z满足(x,y,z)=1,那么这个毕达哥拉斯三元组称为本原的。
    [定理]:正整数x,y,z构成一个本原毕达哥拉斯三元组且y为偶数，当且仅当互素的正整数m,n(m>n),其中m为奇数n为偶数，或者m为偶数n为奇数，并且满足
                   x=m^2-n^2;
                   y=2*m*n;
                   z=m^2+n^2;
    那么要求给定的范围内的本原毕达哥拉斯三元组数，只需对m,n分别枚举即可，然后将三元组乘以i(保证i(x+y+z)在给定的范围之内)倍，就可以求出所有满足条件的毕达哥拉斯三元组。

 【例1】 Right-angled Triangle  [FZU 1669]  点击打开链接

Problem Description

A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments. A triangle with vertices A, B, and C is denoted △ABC.

Triangles can also be classified according to their internal angles, described below using degrees of arc:

• A right triangle (or right-angled triangle, formerly called a rectangled triangle) has one 90° internal angle (a right angle). The side opposite to the right angle is the hypotenuse; it is the longest side in the right triangle. The other two sides are the legs or catheti (singular: cathetus) of the triangle. Right triangles conform to the Pythagorean Theorem, wherein the sum of the squares of the two legs is equal to the square of the hypotenuse, i.e., a^2 + b^2 = c^2, where a and b are the legs and c is the hypotenuse.
• An oblique triangle has no internal angle equal to 90°.
• An obtuse triangle is an oblique triangle with one internal angle larger than 90° (an obtuse angle).
• An acute triangle is an oblique triangle with internal angles all smaller than 90° (three acute angles). An equilateral triangle is an acute triangle, but not all acute triangles are equilateral triangles.

What we consider here is very simple. Give you the length of L, you should calculate there are how many right-angled triangles such that a + b + c ≤ L where a and b are the legs and c is the hypotenuse. You should note that the three numbers a, b and c are all integers.

Input

Output

Sample Input

Sample Output

Hint

There are five right-angled triangles where a + b + c ≤ 40. That are one right-angled triangle where a = 3, b = 4 and c = 5; one right-angled triangle where a = 6, b = 8 and c = 10; one right-angled triangle where a = 5, b = 12 and c = 13; one right-angled triangle where a = 9, b = 12 and c = 15; one right-angled triangle where a = 8, b = 15 and c = 17.  

#include <stdio.h>
#include <string.h>
#include <math.h>long long gcd(long long a,long long b)
{return b==0?a:gcd(b,a%b);
}long long fun(long long a)
{long long i,j,k,tmp,x,y,z,sum=0;tmp=sqrt(a+0.0);for(i=1;i<=tmp;i++){for(j=i+1;j<=tmp;j++){if(2*j*j+2*i*j>a)break;if(i%2!=j%2){if(gcd(i,j)==1){x=j*j-i*i;y=2*i*j;z=i*i+j*j;for(k=1;;k++){if(k*(x+y+z)>a)break;sum++;}}}}}return sum;
}
int main()
{long long n;while(scanf("%lld",&n)!=EOF){long long sum=fun(n);printf("%lld\n",sum);}
}