如果x^2+y^2=r^2，则(r-x)(r+x)=y*y
令d=gcd(r-x,r+x),r-x=d*u^2,r+x=d*v^2,显然有gcd(u,v)=1且u<v
有2r=d*(u^2+v^2)，y=d*u*v，x=d(v^2-u^2)/2
枚举2r的约数d，再花费sqrt(2r/d)的时间枚举u，求出v=sqrt(2r/d-u^2)然后判断gcd(u,v)=1
最后结果乘以4（四个象限）+4（坐标轴上）即可
/**************************************************************
Problem: 1041
User: ******
Language: C++
Result: Accepted
Time:92 ms
Memory:1284 kb
****************************************************************/
#include <vector>
#include <list>
#include <limits.h>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string.h>
#include <stdlib.h>
#include <cassert>
using namespace std;
int main() {
long long r;
cin >> r;
long long d, a, b;
long long ans = 0;
for (d = 1; d * d <= 2*r; ++d) {
if ((2 * r) % d) continue;
long long t = (2 * r) / d;
for (a = 1; a * a <= t; ++a) {
b = sqrt(t - a * a);
if (b * b != t - a * a) continue;
if (a >= b) continue;
if (__gcd(a, b) > 1) continue;
// cout << d << " " << a << " " << b << endl;
// cout << "X: " << - a * a * d + r << endl;
++ans;
}
if (d * d == 2 * r) continue;
t = d;
for (a = 1; a * a <= t; ++a) {
b = sqrt(t - a * a);
if (b * b != t - a * a) continue;
if (a >= b) continue;
if (__gcd(a, b) > 1) continue;
// cout << t << " " << a << " " << b << endl;
// cout << "X: " << - a * a * (2*r/d) + r << endl;
++ans;
}
}
ans = (ans + 1) * 4;
cout << ans << endl;
return 0;
}
转摘至：http://blog.csdn.net/lwfcgz/article/details/39927801