$$\sum _{i=1}^n(-1{)}^{\lfloor i\times \sqrt{k}\rfloor },(n\le 10^9,k\le 10^4)$$
考虑$(-1{)}^x=1-2\times (x\mathrm{m}\mathrm{o}\mathrm{d}2)=1-2(x-2\times \frac{x}{2})=1-2x+4\times \lfloor \frac{x}{2}\rfloor$。
$$\begin{gathered} \sum _{i=1}^n(-1{)}^{\lfloor i\times \sqrt{k}\rfloor } \\ n-2\times \sum _{i=1}^n\lfloor i\times \sqrt{k}\rfloor +4\times \sum _{i=1}^n\lfloor \frac{i\times \sqrt{k}}{2}\rfloor \end{gathered}$$
我么假设$f(a,b,c,n)=\sum _{i=1}^n\lfloor \frac{\left(a\times \sqrt{r}+b\right)\times i}{c}\rfloor$，

如果$\frac{a\times \sqrt{r}+b}{c}\ge 1,d=\lfloor \frac{a\times \sqrt{r}+b}{c}\rfloor$：
$$\begin{gathered} d\times \frac{n\times (n+1)}{2}+\sum _{i=1}^n\lfloor \frac{\left(a\times \sqrt{r}+b-d\times c\right)\times i}{c}\rfloor \\ f(a,b,c,n)=d\times \frac{n\times (n+1)}{2}+f(a,b-d\times c,c,n) \end{gathered}$$

否则$\frac{a\times \sqrt{r}+b}{c}<1$：

由类欧的一般套路，我们设$m=\lfloor n\times \frac{a\sqrt{r}+b}{c}\rfloor ,t=\frac{a\times \sqrt{r}+b}{c}$，这里的$d$没有向下取整，是一个实数。
$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^m[j<t\times i] \\ \sum _{i=1}^n\sum _{j=1}^m[i>\frac{j}{t}] \\ \sum _{j=1}^{m}n-\lfloor \frac{j}{t}\rfloor \\ \sum _{i=1}^{m}n-\lfloor \frac{i\times c}{a\times \sqrt{r}+b}\rfloor \\ n\times m-\sum _{i=1}^m\lfloor \frac{i\times c}{a\times \sqrt{r}+b}\rfloor \end{gathered}$$
考虑将分母有理化，也就是乘上一个$a\times \sqrt{r}-b$。
$$n\times m-\sum _{i=1}^m\lfloor \frac{ca\sqrt{r}-cb}{a^{2}r-b^2}i\rfloor$$
综上，可以通过类欧来解决，递归出口就是$n=0$。

#include <bits/stdc++.h>using namespace std;double x;int r;long long f(long long a, long long b, long long c, long long n) {if (!n) {return 0;}long long t = (a * x + b) / c;if (t) {return n * (n + 1) / 2 * t + f(a, b - t * c, c, n);}long long A = c * a, B = -c * b, C = a * a * r - b * b;long long d = __gcd(A, __gcd(B, C)), m = n * (a * x + b) / c;A /= d, B /= d, C /= d;return n * m - f(A, B, C, m);
}int main() {// freopen("in.txt", "r", stdin);// freopen("ou.txt", "w", stdout);int T, n;scanf("%d", &T);while (T--) {scanf("%d %d", &n, &r);x = sqrt(r);int s = sqrt(r);if (s * s == r) {if (s % 2 == 0) {printf("%d\n", n);}else if (n % 2 == 0) {puts("0");}else {puts("-1");}}else {printf("%lld\n", n - 2 * f(1, 0, 1, n) + 4 * f(1, 0, 2, n));}}return 0;
}