sgu 136 分类： sgu 2015-03-14 22:38 34人阅读 评论(0) 收藏

构造题

根据中点公式列方程，
然后用代入法手算解方程

最后判断方程的根的个数，
如果有根，就构造一组方案

---

---

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<iostream>
#include<algorithm>

const int MAXN = 10005;
const double eps = 1e-8;

int n;
struct node
{
    double x,y;
    node(double x = 0, double y = 0):x(x), y(y){}
    void in(){scanf("%lf%lf",&x,&y);}
    void out(){printf("%lf %lf\n",x,y);}

}A[MAXN] = {0}, B[MAXN] = {0};

node operator+(const node &a,const node &b){return node(a.x+b.x,a.y+b.y);}
node operator-(const node &a,const node &b){return node(a.x-b.x,a.y-b.y);}
node operator*(const node &a,const int &b) {return node(a.x*b,a.y*b);}


int main()
{
#ifndef ONLINE_JUDGE
     freopen("sgu136.in","r",stdin);
     freopen("sgu136.out","w",stdout);
#endif

     scanf("%d",&n);

     for(int i = 1; i <= n ; i++)B[i].in();

     if(n&1)
     {
        for(int i = 1; i <= n; i++)
           if(i&1)A[1] = A[1] + B[i];
           else   A[1] = A[1] - B[i];    
        for(int i = 2; i <= n ;i++)
           A[i] = B[i-1] * 2 - A[i-1]; 

        puts("YES");
        for(int i = 1; i <= n; i++)A[i].out();  
     }
     else
     {
        for(int i = 1; i <= n; i++)
           if(i&1)A[1] = A[1] + B[i];
           else   A[1] = A[1] - B[i];   

        if(fabs(A[1].x) < eps && fabs(A[1].y)<eps)
        {
           A[1].x = A[1].y = 0.1;   

           for(int i = 2; i <= n ;i++)
              A[i] = B[i-1] * 2 - A[i-1]; 

           puts("YES");
           for(int i = 1; i <= n; i++)A[i].out(); 
        }
        else
            puts("NO");

     }

#ifndef ONLINE_JUDGE
     fclose(stdin);
     fclose(stdout);
#endif     
}

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