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