bzoj 1041 圆上的整点

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这里先只考虑x，y都大于0的情况

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

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