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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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/*
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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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using namespace std;
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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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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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*/
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int pf(int x) {
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return x * x;
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}
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bool check(Node a, Node b, Node c) {
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if (a.c != b.c || a.c != c.c || b.c != c.c) return false;
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// a与b之间距离 不等于 a与c之间距离
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if (pf(a.x - b.x) + pf(a.y - b.y) * 3 != pf(a.x - c.x) + pf(a.y - c.y) * 3) return false;
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// a与b之间距离 不等于 b与c之间距离
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if (pf(a.x - b.x) + pf(a.y - b.y) * 3 != pf(b.x - c.x) + pf(b.y - c.y) * 3) return false;
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return true;
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("ZhiMuSanJiaoXing.in", "r", stdin);
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#endif
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int n;
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cin >> n;
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string s;
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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 = 0; i < n; i++) // n行
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for (int j = 0; j <= i; j++) // i列
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a[i][j] = s[idx++];
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// 把所有坐标记录下来
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for (int i = 0; i < n; i++) {
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for (int j = 0; j <= i; j++) {
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q.push_back({n - 1 - i, j, a[i][j]});
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// cout << "x=" << n - 1 - i << ",y=" << j << " " << a[i][j] << endl;
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}
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}
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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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}
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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 << endl;
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// }
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return 0;
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} |
