实数的线性基配用高斯消元思想食用

其实这道题是高斯消元的题

但n^3也还是可以，要在膜质数下进行

再加点贪心

#include
      
       #define re return#define inc(i,l,r) for(int i=l;i<=r;++i) using namespace std;template
       
        inline void rd(T&x){    char c;bool f=0;    while((c=getchar())<'0'||c>'9')if(c=='-')f=1;    x=c^48;    while((c=getchar())>='0'&&c<='9')x=x*10+(c^48);    if(f)x=-x;}typedef double D;D EPS=1e-4;const int maxn=505;int p[maxn];int cost[maxn],n,m,sum,ans; inline D Fabs(D x){    re x>0?x:-x;}struct node{    D x[maxn];    int cost;    inline bool operator<(node c)const     {        re cost
        
         EPS)    {        if(!p[j])//加入基         {            p[j]=i;            ++sum;            ans+=a[i].cost;            break;        }                D t=a[i].x[j]/a[p[j]].x[j];//消元         inc(k,j,m)        a[i].x[k]-=a[p[j]].x[k]*t;    }        printf("%d %d",sum,ans);    re 0;}