[x^2+y^2=r^2
\\
y^2=left( r+x
ight) left( r-x
ight)
\\
ext{let }d= ext{gcd}left( r+x,r-x
ight)
\\
ext{let }r-x=d imes a^2,r+x=d imes b^2
\\
2r=d imes left( a^2+b^2
ight)
\\
d|2r\,\,a^2ot b^2
\\
]
用到了一个结论,两个互质的数的乘积是完全平方数,那么这两个数都是完全平方数
枚举d再枚举a
inline void solve(uint x) {
for(uint i = 1; 2 * i * i <= x; ++i) {
uint j = sqrt(x - i * i);
if (j * j == x - i * i && gcd(i, j) == 1) ++ans;
}
}
cin >> r;r <<= 1;
for(uint i = 1; i * i <= r; ++i) {
if (r % i) continue;
solve(r / i);
if (i * i != r) solve(i);
}
cout << ans * 4;