From b3aff5ce01ffa004319c7d3e5a27db963613581c Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Fri, 8 Mar 2024 10:24:06 +0800
Subject: [PATCH] 'commit'

---
 .../ZhongGaoJi/LanQiao15STEMA202401/3.cpp     | 44 +++++++++++++++++++
 1 file changed, 44 insertions(+)
 create mode 100644 TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/3.cpp

diff --git a/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/3.cpp b/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/3.cpp
new file mode 100644
index 0000000..f21c4e0
--- /dev/null
+++ b/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/3.cpp
@@ -0,0 +1,44 @@
+#include <bits/stdc++.h>
+using namespace std;
+/*
+测试样例：
+abcdefgh
+
+输出：
+aebfcgdh
+
+思考：
+行数：n, 列数：m 需要满足 m>=n
+(1) 1行， m=length(abcdefgh)=8 n=1 满足m>=n
+(2) 2行， m=length(abcdefgh)/2=8/2=4 n=2 满足m>=n
+(3) 3行， 8不能被3整除，跳过
+(4) 4行， m=8/4=2, n=4 m<n break
+*/
+const int N = 210;
+char a[N][N];
+
+int main() {
+    string s;
+    cin >> s; // 生成一个n行m列的矩阵
+    // 1、整明白到底是几行几列
+    int len = s.size();
+    int n = 0; // 最后是几行
+    for (int i = 1;; i++) {
+        if (len % i == 0 && len / i >= i) n = i;
+        if (len / i < i) break;
+    }
+
+    int m = len / n; // 每行多少列
+
+    // 2、做一个二维矩阵存进去
+    for (int i = 0; i < n; i++)     // 行
+        for (int j = 0; j < m; j++) // 列
+            a[i][j] = s[(i)*m + j];
+
+    // 3、从第一列开始，逐列从上下到输出矩阵中的字符
+    for (int j = 0; j < m; j++)
+        for (int i = 0; i < n; i++)
+            cout << a[i][j];
+
+    return 0;
+}
\ No newline at end of file