Codeforces Round #262 (Div. 2) E. Roland and Rose 暴力

E. Roland and Rose

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/460/E

Description

Roland loves growing flowers. He has recently grown a beautiful rose at point (0, 0) of the Cartesian coordinate system. The rose is so beautiful that Roland is afraid that the evil forces can try and steal it.

To protect the rose, Roland wants to build n watch towers. Let's assume that a tower is a point on the plane at the distance of at most r from the rose. Besides, Roland assumes that the towers should be built at points with integer coordinates and the sum of squares of distances between all pairs of towers must be as large as possible. Note, that Roland may build several towers at the same point, also he may build some of them at point (0, 0).

Help Roland build the towers at the integer points so that the sum of squares of distances between all towers is maximum possible. Note that the distance in this problem is defined as the Euclidian distance between points.

Input

The first line contains two integers, n and r (2 ≤ n ≤ 8; 1 ≤ r ≤ 30).

1000000000.

Output

In the first line print an integer — the maximum possible sum of squared distances. In the i-th of the following n lines print two integers, x_i, y_i — the coordinates of the i-th tower. Each tower must be inside or on the border of the circle with radius r. Note that there may be several towers located at the same point of the plane, also some towers can be located at point (0, 0).

If there are multiple valid optimal arrangements, choose any of them.

Sample Input

4 1

Sample Output

16
0 1
0 1
0 -1
0 -1

HINT

题意

在以原点为圆心，半径为R的局域内选择N个整数点，使得N个点中两两距离的平方和最大。

题解：

R最大为30，那么其实距离圆心距离最大的整数点不过12个最多，直接暴力枚举。

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff;   //无限大
const int inf=0x3f3f3f3f;
/*

int buf[10];
inline void write(int i) {
  int p = 0;if(i == 0) p++;
  else while(i) {buf[p++] = i % 10;i /= 10;}
  for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);
  printf("\n");
}
*/
//**************************************************************************************
inline ll read()
{
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
const int K = , N = ;

int x[K], y[K], z[K], ax[N], ay[N], tx[N], ty[N], n, m, r, i, j, k, r1, r2, s;

void go(int t, int p)
{
    if (p == n)
    {
        int u = ;
        for (int i = ; i < n; i++)
            for (int j = i + ; j < n; j++)
                u += (tx[i] -  tx[j]) * (tx[i] - tx[j]) + (ty[i] - ty[j]) * (ty[i] - ty[j]);
        if (u > s)
        {
            s = u;
            for (int i = ; i < n; i++)
                ax[i] = tx[i], ay[i] = ty[i];
        }
    }
    else
    {
        for (int i = t; i < k; i++)
        {
            tx[p] = x[i], ty[p] = y[i];
            go(i, p + );
        }
    }
}

int main()
{
    scanf("%d%d", &n, &r);

    r1 = (r - ) * (r - );
    r2 = r * r;

    for (i = -r; i <= r; i++)
        for (j = -r; j <= r; j++)
        {
            int t = i * i + j * j;
            if (t <= r2 && t > r1)
                x[k] = i, y[k] = j, z[k++] = t;
        }

    for (i = ; i < k; i++)
        for (j = i + ; j < k; j++)
            if (z[j] > z[i])
            {
                int zz;
                zz = x[j], x[j] = x[i], x[i] = zz;
                zz = y[j], y[j] = y[i], y[i] = zz;
                zz = z[j], z[j] = z[i], z[i] = zz;
            }
    k = ;

    go(, );

    printf("%d\n", s);
    for (i = ; i < n; i++)
        printf("%d %d\n", ax[i], ay[i]);

    return ;
}