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#include <bits/stdc++.h>
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const int N = 110;
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char a[N][N];
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using namespace std;
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const double eps = 1e-8;
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int n;
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struct Node {
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int x, y;
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char c;
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};
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vector<Node> q;
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/*
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4
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abccddadca
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2
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aaa
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*/
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/*
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把半径当做 1 ，建立坐标系，然后枚举就行了，需要注意的是，在判断三条边的长度是否相等时用 double 牵扯到精度问题，
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解决方法是:比较长度的平方，长度的平方肯定是整数，还有就是层层之间纵坐标相差是sqrt(3) 的倍数，
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为了不牵扯到小数，把纵坐标的 sqrt(3) 当做1 ，求边长时乘三即可。
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共n=4行
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第一行第一个3,3*sqrt(3)
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第二行第一个:2,2*sqrt(3) 第二行第二个: 2+2,2*sqrt(3)
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第三行第一个:1,1*sqrt(3) 第三行第二个: 1+2,1*sqrt(3) 第三行第三个:1+2+2,1*sqrt(3)
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第四行第一个:0,0 第四行第二个:0+2,0 第四行第三个:0+2+2,0 第四行第四个:0+2+2+2,0
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规律
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第i行，第j个
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i从1开始，到n=4结束
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j从1开始，到i结束
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找出递推关系式：
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x=n-i + (j-1)*2
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y=(n-i)*sqrt(3)
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*/
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bool check(Node a, Node b, Node c) {
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// x= n - i + (j - 1) * 2
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// y= (n - i) * sqrt(3)
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if (a.c != b.c || a.c != c.c || b.c != c.c) return false;
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double ax = (n - a.x + (a.y - 1) * 2);
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double bx = (n - b.x + (b.y - 1) * 2);
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double cx = (n - c.x + (c.y - 1) * 2);
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double ay = (n - a.x) * sqrt(3);
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double by = (n - b.x) * sqrt(3);
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double cy = (n - c.x) * sqrt(3);
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double c1 = (ax - bx) * (ax - bx) + (ay - by) * (ay - by);
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double c2 = (ax - cx) * (ax - cx) + (ay - cy) * (ay - cy);
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double c3 = (bx - cx) * (bx - cx) + (by - cy) * (by - cy);
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if (abs(c1 - c2) < eps && abs(c1 - c3) < eps && abs(c2 - c3) < eps) return true;
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return false;
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("51nod_1909_2_in.txt", "r", stdin);
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#endif
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cin >> n; // 4
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string s; // abccddadca
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cin >> s;
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// 先把字符串存入char[][]
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int idx = 0;
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for (int i = 1; i <= n; i++) // n行
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for (int j = 1; j <= i; j++) // i列
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a[i][j] = s[idx++];
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// 把所有序号记录下来
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= i; j++)
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q.push_back({i, j, a[i][j]});
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vector<char> res;
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// 在q数组中，选择任意三个，计算两两间距离是不是相等
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for (int i = 0; i < q.size(); i++)
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for (int j = i + 1; j < q.size(); j++)
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for (int k = j + 1; k < q.size(); k++)
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if (check(q[i], q[j], q[k])) res.push_back(q[i].c);
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if (res.size() == 0)
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cout << "No Solution" << endl;
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else {
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sort(res.begin(), res.end());
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for (char x : res) cout << x;
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}
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return 0;
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} |
