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 = 10007;
const double PI = acos(-1.0);
const double EPS = 1e-7;

struct point
{
    double x, y;
    point(double x=0, double y=0):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=1; i<=N; i++)
    {
        scanf("%lf%lf", &Mid[i].x, &Mid[i].y);
        if(i&1)poly[1] = poly[1]+Mid[i];
        else poly[1] = poly[1]-Mid[i];
    }

    for(int i=1; i<=N; i++)
        poly[i+1] = Mid[i] * 2 - poly[i];

    if(poly[N+1].x != poly[1].x || poly[N+1].y != poly[1].y)
        printf("NO\n");
    else
    {
        printf("YES\n");
        for(int i=1; i<=N; i++)
            printf("%.3f %.3f\n", poly[i].x, poly[i].y);
    }

    return 0;
}