[洛谷P5174]圆点

题目大意：给你$R(Rleqslant10^{14})$，求：
$$
sumlimits_{xinmathbb{Z}}sumlimits_{yinmathbb{Z}}[x^2+y^2leqslant R](x^2+y^2)
$$
题解：明显可以发现这是对称的，所以可以只枚举四分之一，并且$x,yleqslantsqrt R$。所以式子成了$4sumlimits_{x=0}^{sqrt R}sumlimits_{y=1}^{sqrt R}[x^2+y^2leqslant R](x^2+y^2)$。这样就可以暴力枚举了，复杂度$O(R)$。

然后那个判断有点烦，去掉，变成$4sumlimits_{x=0}^{sqrt R}sumlimits_{y=1}^{sqrt{R-x^2}}(x^2+y^2)$
$$
egin{align*}
&4sumlimits_{x=0}^{sqrt R}sumlimits_{y=1}^{sqrt{R-x^2}}(x^2+y^2)\
=&4sumlimits_{x=0}^{sqrt R}(sqrt{R-x^2}x^2+sumlimits_{y=1}^{sqrt{R-x^2}}y^2)\
end{align*}\
令Y=sqrt{R-x^2}\
=4sumlimits_{x=0}^{sqrt R}(Yx^2+dfrac{Y(Y+1)(2Y+1))}6)\
$$

复杂度$O(sqrt R)$

卡点：无

C++ Code：

#include <cstdio>
#include <cmath>
const long long mod = 1e9 + 7, mod6 = mod * 6;
long long R, ans;
inline void reduce(long long &x) { x += x >> 63 & mod; }
int main() {
	scanf("%lld", &R);
	for (long long x = sqrt(R), y; ~x; --x) {
		y = sqrt(R - x * x);
		reduce(ans += x * x % mod * y % mod - mod);
		reduce(ans += y * (y + 1) % mod6 * (2 * y + 1) / 6 % mod - mod);
	}
	printf("%lld
", ans * 4 % mod);
	return 0;
}