HDU 5365

Run

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 957    Accepted Submission(s): 413

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)

 

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/*
 * In the name of god (^_^)
 */
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
typedef pair<int , int> pii;
#define maxn 22

int n;

struct point
{
    int x, y;
} P[maxn];

int dist(point a, point b)
{
    return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}

int D[6];
LL ans ;

int main()
{
    while(~scanf("%d", &n))
    {
        ans = 0;
        for(int i = 1; i <= n; i++)
            scanf("%d%d", &P[i].x, &P[i].y);

        for(int i = 1; i + 3 <= n; i++)
            for(int j = i + 1; j + 2 <= n; j++)
                for(int t = j + 1; t + 1 <= n; t++)
                    for(int k = t + 1; k <= n; k++)
                    {
                        D[0] = dist(P[i], P[j]);
                        D[1] = dist(P[i], P[t]);
                        D[2] = dist(P[i], P[k]);
                        D[3] = dist(P[j], P[t]);
                        D[4] = dist(P[j], P[k]);
                        D[5] = dist(P[t], P[k]);
                        sort(D, D + 6);
                        int num;
                        for(num = 0; num < 3; num++)
                        {
                            if(D[num] != D[num + 1])
                            break;
                        }
                        if(num == 3 && D[num + 1] == 2 * D[num] && D[num])
                        {
                            ans++;
                        }
                    }
        printf("%I64d\n", ans);
    }
    return 0;
}


/*

4
0 0
0 0
0 0
0 0

5
0 1
1 0
1 2
2 1
2 2

1
1 1

7
0 0
0 1
1 0
1 1
0 2
2 0
2 2

1
0 0

*/