标签:mes scan clu ble namespace image include apr ges
题意:让你构造一个n个点的简单多边形,使得所有点是整点,并且所有边长是整数,并且没有边平行于坐标轴。
就利用勾股数,如下图这样构造即可,n为偶数时,只需矩形拼成,n为奇数时,封上虚线边即可。
#include<cstdio> using namespace std; struct Point{ int x,y; Point(const int &x,const int &y){ this->x=x; this->y=y; } Point(){} }p[1005]; int n,e; int main(){ // freopen("k.in","r",stdin); bool tag=0; int x=-8,y=-19; scanf("%d",&n); if(n==3){ puts("0 0\n4 3\n-20 21"); return 0; } if(n%2==1){ ++n; tag=1; } p[++e]=Point(0,0); p[++e]=Point(-12,-16); p[++e]=Point(-8,-19); p[++e]=Point(4,-3); for(int i=1;i<=(n-4)/2;++i){ if(i%2==1){ Point a[3]; a[0]=Point(x+4*4,y-4*3); a[1]=Point(a[0].x+3,a[0].y+4); a[2]=Point(x+3,y+4); for(int j=e;j>=e/2+2;--j){ p[j+2]=p[j]; } p[e/2+1]=a[0]; p[e/2+2]=a[1]; p[e/2+3]=a[2]; x=a[0].x; y=a[0].y; e+=2; } else{ Point a[3]; a[2]=Point(x-3,y-4); a[1]=Point(a[2].x-4*4,a[2].y+4*3); a[0]=Point(a[1].x+3,a[1].y+4); for(int j=e;j>=e/2+1;--j){ p[j+2]=p[j]; } p[e/2]=a[0]; p[e/2+1]=a[1]; p[e/2+2]=a[2]; x=a[2].x; y=a[2].y; e+=2; } } if(tag){ p[e-1]=p[e]; --e; } for(int i=1;i<=e;++i){ printf("%d %d\n",p[i].x,p[i].y); } return 0; }
【构造】Ural Championship April 30, 2017 Problem K. King’s island
标签:mes scan clu ble namespace image include apr ges
原文地址:http://www.cnblogs.com/autsky-jadek/p/7618368.html