复数模版
写着玩的,以证明我这两天没颓废……= =
大约是本blog最低水平的一篇???
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cmath> #include<cstdio> #include<algorithm> #define eps 1e-9 #define ld long double using namespace std; int a,b,c,d,e; struct com{ ld r,i; com(ld _r=0,ld _i=0):r(_r),i(_i){}; void pr(){printf("%Lf %Lf\n",r,i);} }s[4],w; com operator + (com a,com b){return com(a.r+b.r,a.i+b.i);} com operator - (com a,com b){return com(a.r-b.r,a.i-b.i);} com operator * (com a,com b){return com(a.r*b.r-a.i*b.i,a.r*b.i+a.i*b.r);} com operator / (com a,com b){ld c=b.r*b.r+b.i*b.i;return com((a.r*b.r+a.i*b.i)/c,(a.i*b.r-a.r*b.i)/c);} bool operator < (com a,com b){return a.r<b.r&&a.i<b.i;} bool operator == (com a,com b){return fabs(a.r-b.r)<eps&&fabs(a.i-b.i)<eps;} com sqrt(com a){ if (fabs(a.i)<=eps) if (a.r>0) return com(sqrt(a.r),0);else return com(0,sqrt(-a.r)); com b; b.i=sqrt((sqrt(a.r*a.r+a.i*a.i)-a.r)/2); b.r=a.i/2/b.i; return b; } struct jie{com x1,x2;jie(com _x1=0,com _x2=0):x1(_x1),x2(_x2){};}; inline jie work(com a,com b,com c){ com det=sqrt(b*b-4*a*c); return jie((det-b)/com(2,0)/a,(0-det-b)/com(2,0)/a); } int main(){ }