Regular polygon
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2004 Accepted Submission(s): 795
Problem Description
On
a two-dimensional plane, give you n integer points. Your task is to
figure out how many different regular polygon these points can make.
Input
The
input file consists of several test cases. Each case the first line is a
numbers N (N <= 500). The next N lines ,each line contain two number
Xi and Yi(-100 <= xi,yi <= 100), means the points’ position.(the
data assures no two points share the same position.)
Output
For each case, output a number means how many different regular polygon these points can make.
Sample Input
4
0 0
0 1
1 0
1 1
6
0 0
0 1
1 0
1 1
2 0
2 1
Sample Output
1
2
题意：给你n个整数点 问你组成正多边形的有几组
因为题目给的是整数点 只要正四边形才能是整点
所以我们只要枚举其中任意两个点 利用旋转公式可以得到其它两个点 判断其他两个点是否存在就可以了
旋转公式是 x=xcos@-ysin@
y=xsin@-ycos@;
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #include<cstdlib> 6 #include<string.h> 7 #include<set> 8 #include<vector> 9 #include<queue> 10 #include<stack> 11 #include<map> 12 #include<cmath> 13 typedef long long ll; 14 typedef unsigned long long LL; 15 using namespace std; 16 const double PI=acos(-1.0); 17 const double eps=0.0000000001; 18 const int N=1000+100; 19 struct node{ 20 int x,y; 21 }a[N]; 22 int vis[N][N]; 23 int main(){ 24 int n; 25 while(scanf("%d",&n)!=EOF){ 26 memset(vis,0,sizeof(vis)); 27 for(int i=1;i<=n;i++){ 28 scanf("%d%d",&a[i].x,&a[i].y); 29 vis[a[i].x+500][a[i].y+500]=i; 30 } 31 ll ans=0; 32 int x,y; 33 int xx,yy; 34 int len; 35 for(int i=1;i<=n;i++){ 36 for(int j=i+1;j<=n;j++){ 37 if(i==j)continue; 38 x=(a[j].y-a[i].y)+a[i].x; 39 y=(a[j].x-a[i].x)*(-1)+a[i].y; 40 xx=-(a[i].y-a[j].y)+a[j].x; 41 yy=(a[i].x-a[j].x)+a[j].y; 42 int t1=vis[x+500][y+500]; 43 int t2=vis[xx+500][yy+500]; 44 if(t1&&t2)ans++; 45 x=-(a[j].y-a[i].y)+a[i].x; 46 y=(a[j].x-a[i].x)+a[i].y; 47 xx=(a[i].y-a[j].y)+a[j].x; 48 yy=-(a[i].x-a[j].x)+a[j].y; 49 int t3=vis[x+500][y+500]; 50 int t4=vis[xx+500][yy+500]; 51 if(t3&&t4)ans++; 52 } 53 } 54 // cout<<ans<<endl; 55 ans=ans/4; 56 printf("%I64d ",ans); 57 } 58 }