题目链接：Codeforces 460E Roland and Rose

题目大意：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。

解题思路：R最大为30。那么事实上距离圆心距离最大的整数点只是12个最多，直接暴力枚举。

#include <cstdio>

#include <cstring>

#include <vector>

#include <algorithm>

using namespace std;

struct point {

    int x, y;

    point (int x = 0, int y = 0) {

        this->x = x;

        this->y = y;

    }

};

int N, R, M, ans, pos[10], rec[10];

vector<point> vec;

inline int dis (int x, int y) {

    return x * x + y * y;

}

inline bool cmp (const point& a, const point& b) {

    return dis(a.x, a.y) > dis(b.x, b.y);

}

void init () {

    scanf("%d%d", &N, &R);

    for (int i = -R; i <= R; i++) {

        for (int j = -R; j <= R; j++)  {

            if (i * i + j * j <= R * R)

                vec.push_back(point(i, j));

        }

    }

    ans = 0;

    M = min((int)vec.size(), 18);

    sort(vec.begin(), vec.end(), cmp);

}

void dfs (int d, int f, int s) {

    if (d == N) {

        if (s > ans) {

            ans = s;

            memcpy(rec, pos, sizeof(pos));

        }

        return;

    }

    for (int i = f; i < M; i++) {

        int add = 0;

        for (int j = 0; j < d; j++)

            add += dis(vec[pos[j]].x - vec[i].x, vec[pos[j]].y - vec[i].y);

        pos[d] = i;

        dfs(d + 1, i, s + add);

    }

}

int main () {

    init();

    dfs(0, 0, 0);

    printf("%d\n", ans);

    for (int i = 0; i < N; i++)

        printf("%d %d\n", vec[rec[i]].x, vec[rec[i]].y);

    return 0;

}