diff --git a/TangDou/LuoGuBook/ZhiMuSanJiaoXing.cpp b/TangDou/LuoGuBook/ZhiMuSanJiaoXing.cpp
index 1762eba..a736d88 100644
--- a/TangDou/LuoGuBook/ZhiMuSanJiaoXing.cpp
+++ b/TangDou/LuoGuBook/ZhiMuSanJiaoXing.cpp
@@ -1,74 +1,94 @@
 #include <bits/stdc++.h>
 const int N = 110;
 char a[N][N];
-/*
-4
-abccddadca
 
-2
-aaa
-*/
 using namespace std;
-
+const double eps = 1e-8;
+int n;
 struct Node {
     int x, y;
     char c;
 };
 vector<Node> q;
+
+/*
+4
+abccddadca
+
+2
+aaa
+*/
 /*
 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，
 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，
 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。
+
+
+共n=4行
+第一行第一个3,3*sqrt(3)
+第二行第一个:2,2*sqrt(3)   第二行第二个: 2+2,2*sqrt(3)
+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)
+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
+
+规律
+第i行，第j个
+i从1开始，到n=4结束
+j从1开始，到i结束
+x=n-i + (j-1)*2
+y=(n-i)*sqrt(3)
+
 */
-int pf(int x) {
-    return x * x;
-}
 
 bool check(Node a, Node b, Node c) {
+    //  n - i + (j - 1) * 2, (n - i) * sqrt(3)
     if (a.c != b.c || a.c != c.c || b.c != c.c) return false;
-    // a与b之间距离 不等于 a与c之间距离
-    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;
-    // a与b之间距离 不等于 b与c之间距离
-    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;
 
-    return true;
+    double ax = (n - a.x + (a.y - 1) * 2);
+    double bx = (n - b.x + (b.y - 1) * 2);
+    double cx = (n - c.x + (c.y - 1) * 2);
+
+    double ay = (n - a.x) * sqrt(3);
+    double by = (n - b.x) * sqrt(3);
+    double cy = (n - c.x) * sqrt(3);
+
+    double c1 = (ax - bx) * (ax - bx) + (ay - by) * (ay - by);
+    double c2 = (ax - cx) * (ax - cx) + (ay - cy) * (ay - cy);
+    double c3 = (bx - cx) * (bx - cx) + (by - cy) * (by - cy);
+    if (abs(c1 - c2) < eps && abs(c1 - c3) < eps && abs(c2 - c3) < eps) return true;
+    return false;
 }
 
 int main() {
 #ifndef ONLINE_JUDGE
     freopen("ZhiMuSanJiaoXing.in", "r", stdin);
 #endif
-    int n;
+
     cin >> n;
     string s;
     cin >> s;
     // 先把字符串存入char[][]
     int idx = 0;
-    for (int i = 0; i < n; i++)      // n行
-        for (int j = 0; j <= i; j++) // i列
+    for (int i = 1; i <= n; i++)     // n行
+        for (int j = 1; j <= i; j++) // i列
             a[i][j] = s[idx++];
 
-    // 把所有坐标记录下来
-    for (int i = 0; i < n; i++) {
-        for (int j = 0; j <= i; j++) {
-            q.push_back({n - 1 - i, j, a[i][j]});
-            // cout << "x=" << n - 1 - i << ",y=" << j << " " << a[i][j] << endl;
-        }
-    }
+    // 把所有序号记录下来
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= i; j++)
+            q.push_back({i, j, a[i][j]});
 
     vector<char> res;
     // 在q数组中，选择任意三个，计算两两间距离是不是相等
     for (int i = 0; i < q.size(); i++)
         for (int j = i + 1; j < q.size(); j++)
-            for (int k = j + 1; k < q.size(); k++) {
+            for (int k = j + 1; k < q.size(); k++)
                 if (check(q[i], q[j], q[k])) res.push_back(q[i].c);
-            }
-
-    // if (res.size() == 0)
-    //     cout << "No Solution" << endl;
-    // else {
-    //     sort(res.begin(), res.end());
-    //     for (char x : res) cout << x << endl;
-    // }
+
+    if (res.size() == 0)
+        cout << "No Solution" << endl;
+    else {
+        sort(res.begin(), res.end());
+        for (char x : res) cout << x << endl;
+    }
     return 0;
 }
\ No newline at end of file