Run

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

AFA is a girl who like runing.Today,he download an app about runing .The app can record the trace of her runing.AFA will start runing in the park.There are many chairs in the park,and AFA will start his runing in a chair and end in
this chair.Between two chairs,she running in a line.she want the the trace can be a regular triangle or a square or a regular pentagon or a regular hexagon.

Please tell her how many ways can her find.

Two ways are same if the set of chair that they contains are same.

Input

There are multiply case.

In each case,there is a integer n(1 < = n < = 20)in a line.

In next n lines,there are two integers xi,yi(0 < = xi,yi < 9) in each line.

Output

Output the number of ways.

Sample Input

4
0 0
0 1
1 0
1 1

Sample Output

1

Source

BestCoder Round #50 (div.2)

Recommend

题目说的有点不清，其实就是要求正方形的个数就可以了。由于整点无法形成正三角形 正五边形 正六边形。判定正方形，只需要，四边相等，对角线相等，也对角线是边的2倍就可以了。

#define N 205
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,xx[N],yy[N];
bool isRec(int a,int b,int c,int d){
    int ll[7],len = 0,e[4];
    e[0] = a;e[1] = b;e[2] = c;e[3] = d;
    for(int i = 0;i<4;i++){
        for(int j = i + 1;j<4;j++){
            ll[len++] = (xx[e[i]] - xx[e[j]]) * (xx[e[i]] - xx[e[j]]) +
                        (yy[e[i]] - yy[e[j]]) * (yy[e[i]] - yy[e[j]]);
        }
    }
    sort(ll,ll+len);
    if(ll[0] == ll[1] && ll[1] == ll[2] && ll[2] == ll[3] && ll[4] == 2 * ll[3]&& ll[4] == ll[5])
        return true;
    return false;
}
int main()
{
     while(S(n)!=EOF)
    {
        FI(n){
            S2(xx[i],yy[i]);
        }
        int ans = 0;
        for(int i =0;i<n;i++){
            for(int j = i+1;j<n;j++){
                for(int k = j+1;k<n;k++){
                    for(int s = k+1;s<n;s++){
                            if(isRec(i,j,k,s)){
                                ans++;
                            }
                    }
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}