Erasing Edges - SGU 136(构造多边形)

题目大意：已知一个多边形上的每条边的中点，还原出来一个多边形。

分析：因为偶数是不固定的，所以可以为任意起点，奇数只有一个，可以所有中点加减算出来第一个点，然后就是简单的向量计算点的位置了......

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#include<stdio.h>

#include<string.h>

#include<math.h>

#include<algorithm>

using namespace std;

const int MAXN = ;

const double PI = acos(-1.0);

const double EPS = 1e-;

struct point

{

    double x, y;

    point(double x=, double y=):x(x),y(y){}

    point operator - (const point &tmp)const{

        return point(x-tmp.x, y-tmp.y);

    }

    point operator + (const point &tmp)const{

        return point(x+tmp.x, y+tmp.y);

    }

    point operator * (const int &t)const{

        return point(x*t, y*t);

    }

}Mid[MAXN], poly[MAXN];

int main()

{

    int N;

    scanf("%d", &N);

    for(int i=; i<=N; i++)

    {

        scanf("%lf%lf", &Mid[i].x, &Mid[i].y);

        if(i&)poly[] = poly[]+Mid[i];

        else poly[] = poly[]-Mid[i];

    }

    for(int i=; i<=N; i++)

        poly[i+] = Mid[i] *  - poly[i];

    if(poly[N+].x != poly[].x || poly[N+].y != poly[].y)

        printf("NO\n");

    else

    {

        printf("YES\n");

        for(int i=; i<=N; i++)

            printf("%.3f %.3f\n", poly[i].x, poly[i].y);

    }

    return ;

}