bzoj1041: [HAOI2008]圆上的整点

题面在这里
emmmm先来吐槽一下下~
%一发dsy上0ms的大佬们。。。
我就比较弱辣要200+ms，，QwQ
做法：
来做一些鬼畜的变形= =
r2=x2+y2
y2=r2−x2
y2=(r+x)(r−x)
y=(r+x)(r−x)−−−−−−−−−−−√
设d=gcd(r+x,r−x)
a=r+xd，b=r−xd
则y=d2×a×b−−−−−−−−√=da×b−−−−√
因为a和b互质，所以a和b是完全平方数
设a=A2,b=B2
则A2=a=r+xd,B2=b=r−xd
两式相加得2rd=A2+B2
所以我们可以考虑枚举d，再枚举A，计算出B，然后判断是否合法，即满足：1) B是整数，2) gcd(A,B)=1。
至此，我们求出某一象限内的个数，因为要四个象限，于是答案×4。
再加上坐标轴的4个点，于是答案再+4。

/*************************************************************    Problem: bzoj 1041 [HAOI2008]圆上的整点    User: fengyuan    Language: C++    Result: Accepted    Time: 276 ms    Memory: 1300 kb    Submit_Time: 2017-12-14 19:52:42*************************************************************/#include<bits/stdc++.h>#define rep(i, x, y) for (int i = (x); i <= (y); i ++)#define down(i, x, y) for (int i = (x); i >= (y); i --)#define mid ((l+r)/2)#define lc (o<<1)#define rc (o<<1|1)#define pb push_back#define mp make_pair#define PII pair<int, int>#define F first#define S second#define B begin()#define E end()using namespace std;typedef long long LL;//headLL r;inline LL gcd(LL a, LL b){ return !b ? a : gcd(b, a%b); }int main(){    scanf("%lld", &r); LL ans = 0;    rep(i, 1, floor(sqrt(2*r))) if (2*r % i == 0){        LL d = i;        rep(j, 1, floor(sqrt(2*r/d))){            LL a = j;            LL b = sqrt(2*r/d-a*a);            if (a*a + b*b != 2*r/d || gcd(a, b) != 1 || b <= a) continue;            ans ++;        }        if (d*d == 2*r || d == 1) continue;        d = 2*r/d;        rep(j, 1, floor(sqrt(2*r/d))){            LL a = j;            LL b = sqrt(2*r/d-a*a);            if (a*a + b*b != 2*r/d || gcd(a, b) != 1 || b <= a) continue;            ans ++;        }    }    printf("%lld\n", ans*4+4);    return 0;}