用c语言代码x^2-n*y^2=1,佩尔方程，寻找非平凡解（即y！=0），输入不同的n

#include <stdio.h>
#include <math.h>

int main()
{
    int n, x, y, a, b, c, d, e, f, g, h;
    double sqrt_n;

    printf("请输入n的值：");
    scanf("%d", &n);

    sqrt_n = sqrt(n);
    a = (int)sqrt_n;
    x = a; y = 1;
    b = n - a * a;

    while (b != 1) {
        c = (int)(sqrt(n) + x) / b;
        d = c * b - x;
        e = b;
        x = b;
        y = d;
        b = e - d * d / b;
    }

    f = x * y + 1;
    g = x;
    h = y;

    while (g * g - n * h * h != 1) {
        g += n * h;
        h = (f - n * g) / h;
        f = (int)((sqrt_n + g) / h) * h - g;
    }

    printf("一组非平凡解为：x=%d, y=%d\n", g, h);

    return 0;
}

注意：该代码只能找到一组非平凡解，如果要找到所有非平凡解，需要进行更多的计算。