[模板]二次剩餘
阿新 • • 發佈:2020-07-22
轉載自https://blog.csdn.net/new_ke_2014/article/details/21829921
#include <iostream> using namespace std; #define LL __int64 LL w; struct Point//x + y*sqrt(w) { LL x; LL y; }; LL mod(LL a, LL p) { a %= p; if (a < 0) { a += p; } return a; } Point point_mul(Point a, Point b, LL p) { Point res; res.xView Code= mod(a.x * b.x, p); res.x += mod(w * mod(a.y * b.y, p), p); res.x = mod(res.x, p); res.y = mod(a.x * b.y, p); res.y += mod(a.y * b.x, p); res.y = mod(res.y , p); return res; } Point power(Point a, LL b, LL p) { Point res; res.x = 1; res.y = 0; while(b) {if (b & 1) { res = point_mul(res, a, p); } a = point_mul(a, a, p); b = b >> 1; } return res; } LL quick_power(LL a, LL b, LL p)//(a^b)%p { LL res = 1; while(b) { if (b & 1) { res = (res * a) % p; } a= (a * a) % p; b = b >> 1; } return res; } LL Legendre(LL a, LL p) // a^((p-1)/2) { return quick_power(a, (p - 1) >> 1, p); } LL equation_solve(LL b, LL p)//求解x^2=b(%p)方程解 { if ((Legendre(b, p) + 1) % p == 0) { return -1;//表示沒有解 } LL a, t; while(true) { a = rand() % p; t = a * a - b; t = mod(t, p); if ((Legendre(t, p) + 1) % p == 0) { break; } } w = t; Point temp, res; temp.x = a; temp.y = 1; res = power(temp, (p + 1) >> 1, p); return res.x; } int main() { LL b, p; scanf("%I64d %I64d", &b, &p); printf("%I64d\n", equation_solve(b, p));//輸出為-1表示,不存在解 return 0; }